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CHAPTER CONNECTIONS
Polynomial and
Rational Functions
CHAPTER OUTLINE
4.1 Synthetic Division; the Remainder
and Factor Theorems 308
4.2 The Zeroes of Polynomial Functions 320
4.3 Graphing Polynomial Functions 337
4.4 Graphing Rational Functions 356
4.5 Additional Insights into Rational Functions 371
4.6 Polynomial and Rational Inequalities 385
In a study of demographics, the population
density of a city and its sur rounding area is
measured using a unit called people per
square mile . The population density is much
greater near the city’s center , and tends to
decrease as you move out into suburban and
rural areas. The density can be modeled using
ax
the for mula D1x2 2
, where D (x ) represents
x b
the density at a distance of x mi from the
center of a city , and a and b are constants
related to a par ticular city and its sprawl. Using
this equation, city planners can deter mine
how far fr om the city’s center the population
drops below a cer tain level, and answer other
impor tant questions to help plan for
future growth.
䊳
This application appears as Exer cise 63 in
Section 4.4.
In Chapters 2 and 3, we were able to find the maximum or minimum value of a parabola and other elementary
functions. In a calculus course, certain techniques are developed that enable us to find these extreme values for
Connections almost any function. The ability to graph functions will play an important role in this development, as well as
to Calculus the skills necessary to solve a variety of equation types. These are explored in the Connections to Calculus
feature for Chapter 4.
307
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Synthetic Division; the Remainder and Factor Theorems
LEARNING OBJECTIVES
To find the zero of a linear function, we can use properties of equality to isolate x. To
find the zeroes of a quadratic function, we can factor or use the quadratic formula. To
find the zeroes of higher degree polynomials, we must first develop additional tools,
including synthetic division and the remainder and factor theorems. These will help us
write a higher degree polynomial in terms of linear and quadratic polynomials, whose
zeroes can easily be found.
In Section 4.1 you will see
how we can:
A. Divide polynomials
using long division and
synthetic division
B. Use the remainder
theorem to evaluate
polynomials
C. Use the factor theorem
to factor and build
polynomials
D. Solve applications using
the remainder theorem
EXAMPLE 1
A. Long Division and Synthetic Division
To help understand synthetic division and its use as a mathematical tool, we first
review the process of long division.
Long Division
Polynomial long division closely resembles the division of whole numbers, with the
main difference being that we group each partial product in parentheses to prevent
errors in subtraction.
䊳
Dividing Polynomials Using Long Division
Divide x3 4x2 x 6 by x 1.
Solution
䊳
The divisor is 1x 12 and the dividend is 1x3 4x2 x 62. To find the first
multiplier, we compute the ratio of leading terms from each expression. Here the
x3 from dividend
ratio
shows our first multiplier will be “x2,” with x2 1x ⴚ 12 ⴝ x3 ⴚ x2.
x from divisor
x2
x ⴚ 1冄 x 4x2 x 6
x2
x ⴚ 1冄 x 4x2 x 6
3
1x3 x2 2
subtraction
3
S
x3 x2 algebraic addition
ⴚ3x2 x
At each stage, after writing the subtraction as algebraic addition (distributing
the negative) we compute the sum in each column and “bring down” the next term.
axk next leading term
Each following multiplier is found as before, using the ratio
.
x from divisor
x2 3x 2
x 1冄 x 4x2 x 6
1x3 x2 2
ⴚ3x2 x
next multiplier: ⴚ3x
3x
x
13x2 3x2
ⴚ2x 6
next multiplier: ⴚ2x
x 2
12x 22
4
3
2
3x 1x 12 3x 2 3x, subtract 3x 2 3x
algebraic addition, bring down next term
21x 12 2x 2, subtract 2x 2
algebraic addition, remainder is 4
x 4x x 6
4
x2 3x 2 , or after multiplying
x1
x1
both sides by x 1, x3 4x2 x 6 1x 12 1x2 3x 22 4.
3
2
The result shows
Now try Exercises 7 through 12
308
䊳
4–2
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Section 4.1 Synthetic Division; the Remainder and Factor Theorems
The process illustrated is called the division algorithm, and like the division of
whole numbers, the final result can be checked by multiplication.
dividend
divisor
quotient
remainder
check: x3 4x2 x 6 1x 12 1x2 3x 22 4
1x3 3x2 2x x2 3x 22 4
#
divisor quotient
1x3 4x2 x 22 4
combine like terms
x 4x x 6 ✓
3
2
add remainder
In general, the division algorithm for polynomials says
Division of Polynomials
Given polynomials p1x2 and d1x2 0, there exist unique polynomials q(x) and r(x)
such that
p1x2 d1x2q1x2 r1x2,
where r1x2 0 or the degree of r1x2 is less than the degree of d1x2 .
Here, d1x2 is called the divisor, q1x2 is the quotient, and r1x2 is the remainder.
In other words, “a polynomial of greater degree can be divided by a polynomial of
equal or lesser degree to obtain a quotient and a remainder.” As with whole numbers,
if the remainder is zero, the divisor is a factor of the dividend.
Synthetic Division
As the word “synthetic” implies, synthetic division not only simulates the long division process, but also condenses it and makes it more efficient when the divisor is
linear. The process works by capitalizing on the repetition found in the division algorithm. First, the polynomials involved are written in decreasing order of degree, so the
variable part of each term is unnecessary as we can let the position of each coefficient
indicate the degree of the term. For the dividend from Example 1, 1 4 1 6 would
represent the polynomial 1x3 4x2 1x 6. Also, each stage of the algorithm involves a product of the divisor with the next multiplier, followed by a subtraction.
These can likewise be computed using the coefficients only, as the degree of each term
is still determined by its position. Here is the division from Example 1 in the synthetic
division format. Note that we must use the zero of the divisor (as in x 32 for a divisor
of 2x 3, or in this case, “1” from x 1 02 and the coefficients of the dividend in
the following format:
zero of the divisor
1
coefficients of the dividend
4
1
1
6
partial products
↓
The process of synthetic
division is only summarized
here. For a complete discussion,
see Appendix C.
↓
1
coefficients of the quotient
remainder
As this template indicates, the quotient and remainder will be read from the last row.
The arrow indicates we begin by “dropping the leading coefficient into place.” We
then multiply this coefficient by the “divisor,” then place the result in the next column
and add. Note that using the zero of the divisor enables us to add in each column
directly, rather than subtracting then changing to algebraic addition as before.
add
1
1
4
↓
multiply 1 # 1
↓
WORTHY OF NOTE
1
1
3 ↓
1
6
#
multiply divisor coefficient,
place result in next column and add
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In a sense, we “multiply in the diagonal direction,” and “add in the vertical direction.” Repeat the process until the division is complete.
add
1
1
4
1
1
3
6
multiply 1 # 132 3,
↓
1
3
2↓
4
1
6
1
3
2
3
2
↓
place result in next column and add
1
1
↓
add
1
↓
310
4↓
multiply 1 # 122 2,
place result in next column and add
The quotient is read from the last row by noting the remainder is 4, leaving the
coefficients 1 3 2, which translate back into the polynomial x2 3x 2. The
final result is identical to that in Example 1, but the new process is more efficient, since
all stages are actually computed on a single template as shown here:
zero of the divisor
coefficients of the dividend
1
1
4
1
6
↓
1
3
2
1
3
2
coefficients of the quotient
4
remainder
EXAMPLE 2
䊳
Dividing Polynomials Using Synthetic Division
Solution
䊳
Using 2 as our “divisor” (from x 2 02, we set up the synthetic division
template and begin.
Compute the quotient of 1x3 3x2 4x 122 and 1x 22, then check your
answer.
use 2 as a “divisor”
2
1
↓
1
The result shows
Check
䊳
3
2
1
4
2
6
12
12
0
drop lead coefficient into place;
multiply by divisor, place result
in next column and add
x3 3x2 4x 12
x2 x 6, with no remainder.
x2
x3 3x2 4x 12 1x 22 1x2 x 62
1x3 x2 6x 2x2 2x 122
x3 3x2 4x 12 ✓
Now try Exercises 13 through 20
䊳
Note that in synthetic division, the degree of q(x) will always be one less than p(x),
since the process requires a linear divisor (degree 1).
Since the division process is so dependent on the place value (degree) of each
term, polynomials such as 2x3 3x 7, which has no term of degree 2, must be written using a zero placeholder: 2x3 0x2 3x 7. This ensures that like place values
“line up” as we carry out the division.
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Section 4.1 Synthetic Division; the Remainder and Factor Theorems
EXAMPLE 3
䊳
Dividing Polynomials Using a Zero Placeholder
Compute the quotient
Solution
䊳
3
use 3 as a “divisor”
䊳
2x3 3x 7
and check your answer.
x3
2
↓
2
0
6
6
3
18
21
2x3 3x 7 1x 32 12x2 6x 212 70
12x3 6x2 21x 6x2 18x 632 70
2x3 3x 7 ✓
Now try Exercises 21 through 30
Many corporations now pay their
employees monthly to save on
payroll costs. If your monthly salary
was $2037/mo, but you received a
check for only $237, would you
complain? Just as placeholder
zeroes ensure the correct value of
each digit, they also ensure the
correct valuation of each term in
the division process.
䊳
As noted earlier, for synthetic division the divisor must be a linear polynomial and
2x3 3x2 8x 12
,
the zero of this divisor is used. This means for the quotient
2x 3
3
we have 2x 3 0, and x would be used for synthetic division [see
2
Example 6(c)]. Finally, if the divisor is nonlinear, long division must be used.
䊳
Division with a Nonlinear Divisor
Compute the quotient:
Solution
note place holder 0 for “x2” term
2x3 3x 7 12x2 6x 212 1x 32 70
WORTHY OF NOTE
EXAMPLE 4
7
63
70
70
2x3 3x 7
2x2 6x 21 . Multiplying by
x3
x3
The result shows
x 3 gives
Check
311
䊳
2x4 x3 7x2 3
.
x2 2
Write the dividend as 2x4 x3 7x2 0x 3, and the divisor as x2 0x 2.
The quotient of leading terms gives
2x4 from dividend
2x2 as our first multiplier.
2 from divisor
x
2x2 x 3
x 0x 2 冄 2x x 7x2 0x 3
2 2
Multiply 2x 1x 0x 22
12x4 0x3 4x2 2
x3 3x2 0x
2
Multiply x1x 0x 22
1x3 0x2 2x2
2
4
3
3x 2x 3
13x2 0x 62
2x 3
2
Multiply 31x 2 0x 22
subtract (algebraic addition)
bring down next term
subtract (algebraic addition)
bring down next term
subtract (algebraic addition)
remainder is 2x 3
Since the degree of 2x 3 (degree 1) is less than the degree of the divisor (degree 2),
the process is complete.
2x4 x3 7x2 3
2x 3
12x2 x 32 2
x2 2
x 2
Now try Exercises 31 through 34
䊳
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r1x2
p1x2
q1x2 ,
Note that we elected to keep the solution to Example 4 in the form
d1x2
d1x2
instead of multiplying both sides by d(x).
A. You’ve just seen how
we can divide polynomials
using long division and
synthetic division
B. The Remainder Theorem
In Example 2, we saw that 1x3 3x2 4x 122 1x 22 x2 x 6, with
remainder zero. Similar to whole number division, this means x 2 must be a factor of
x3 3x2 4x 12, a fact made clear as we checked our answer: x3 3x2 4x 12 1x 22 1x2 x 62. Now consider the functions p1x2 x3 5x2 2x 8,
p1x2
x3 5x2 2x 8
d1x2 x 3, and their quotient
. Using 3 as the divisor
x3
d1x2
in synthetic division gives
use 3 as a “divisor”
3
1
↓
1
5
3
2
2
6
4
8
12
4
This shows x 3 is not a factor of p(x), since it didn’t divide evenly (the remainder is not zero). However, from the result p1x2 1x 32 1x2 2x 42 4, we make
a remarkable observation — if we evaluate p132, the quotient portion becomes zero,
showing p132 4 (the remainder).
p132 13 32 3 132 2 2132 44 4
Figure 4.1
102 112 4
4
This result can be verified by evaluating p132 in its original form
(also see Figure 4.1):
p1x2 x3 5x2 2x 8
p132 132 3 5132 2 2132 8
27 45 162 8
4
The result is no coincidence, and illustrates the conclusion of the remainder theorem.
The Remainder Theorem
If a polynomial p(x) is divided by 1x c2 using synthetic division,
the remainder is equal to p(c).
This gives us a powerful tool for evaluating polynomials. Where a direct evaluation involves powers of numbers and a long series of calculations, synthetic division
reduces the process to simple products and sums.
EXAMPLE 5
䊳
Using the Remainder Theorem to Evaluate P olynomials
Use the remainder theorem to find p152 for p1x2 x4 3x3 8x2 5x 6.
Verify the result using a substitution.
Solution
䊳
use 5 as a “divisor”
5
1
1
The result shows p152 19.
3
5
2
8
10
2
5
10
5
6
25
19
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Verification using algebra
Verification using technology
p152 152 3152 8152 5152 6
625 375 200 25 6
625 606
19
4
3
2
Now try Exercises 35 through 44
B. You’ve just seen how
we can use the remainder
theorem to evaluate
polynomials
313
䊳
Since p152 19, we know 15, 192 must be a point of the graph of p(x). The ability
to quickly evaluate polynomial functions using the remainder theorem will be used
extensively in the sections that follow.
C. The Factor Theorem
As a consequence of the remainder theorem, when p(x) is divided by x c and the
remainder is 0, p1c2 0 and c is a zero of the polynomial. The relationship between
x c, c, and p1c2 0 are given in the factor theorem.
The Factor Theorem
For a polynomial p(x),
1. If p1c2 0, then x c is a factor of p(x).
2. If x c is a factor of p(x), then p1c2 0.
The remainder and factor theorems often work together to help us find factors of
higher degree polynomials.
EXAMPLE 6
䊳
Using the Factor Theorem to F ind Factors of a Polynomial
Use the factor theorem to determine if
a. x 2
b. x 1
c. 3x 2
are factors of p1x2 3x4 2x3 21x2 32x 12.
Solution
䊳
a. If x 2 is a factor, then p(2) must be 0. Using the remainder theorem we have
2
3
↓
3
2
6
4
21
8
13
32
26
6
12
12
0
Since the remainder is zero, we know p122 0 (remainder theorem) and
1x 22 is a factor (factor theorem).
b. Similarly, if x 1 is a factor, then p112 must be 0.
1
3
↓
3
2
3
5
21
5
16
32
16
48
12
48
60
Since the remainder is not zero, 1x 12 is not a factor of p.
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2
c. The zero of the divisor 3x 2 0 is x , and this value is used in the
3
synthetic division.
2
2
21
12
3
32
3
↓
14
2
0
12
3
0
21
18
0
2
2
Since the remainder is zero, pa b 0 (remainder theorem) and ax b is the
3
3
related factor (factor theorem). The original factor 3x 2 is found by noting
that the quotient polynomial q1x2 3x3 21x 18 has a common factor of
2
three, which will be factored out and applied to x . Starting with the
3
partially factored form we have
2
p1x2 ax b13x3 21x 182
3
2
ax b132 1x3 7x 62
3
13x 22 1x3 7x 62
partially factored form
factor out 3
2
multiply 3ax b
3
This form of simplification will always take place when the zero found using
synthetic division is a fraction and the coefficients of the polynomial are
integers.
Now try Exercises 45 through 56
䊳
As a final note on Example 6, there should be no hesitation to use fractions in the synthetic division process if the given polynomial has integer coefficients. The fraction
will be a zero only if all values in the quotient line are integers (i.e., all of the products
and sums must be integers).
EXAMPLE 7
䊳
Building a Polynomial Using the Factor Theorem
A polynomial p(x) has three zeroes at x 3, 12, and 12. Use the factor
theorem to find the polynomial.
Solution
䊳
Using the factor theorem, the factors of p(x) must be 1x 32, 1x 122, and
1x 222. Computing the product will yield the polynomial.
p1x2 1x 32 1x 122 1x 122
1x 32 1x2 22
x3 3x2 2x 6
Now try Exercises 57 through 64
䊳
Actually, the result obtained in Example 7 is not unique, since any polynomial of
the form a1x3 3x2 2x 62 will also have the same three zeroes for a 僆 ⺢.
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Figure 4.2 shows the graph of Y1 ⫽ p1x2 , as
well as graph of Y2 ⫽ 2p1x2. The only difference
is 2p1x2 has been vertically stretched. Likewise,
the graph of ⫺1p1x2 would be a vertical reflection, but still with the same zeroes. As in previous
graph-to-equation exercises, finding a unique
value for the leading coefficient a requires that
we use a point (x, y) on the graph of p and substitute these values to solve for a. For Example 7,
assume you were also told the graph contains the
point (1, 2), or x ⫽ 1, p112 ⫽ 2. This would
yield:
p1x2 ⫽ a1x3 ⫺ 3x2 ⫺ 2x ⫹ 62
p112 ⫽ a3 1 ⫺ 3112 ⫺ 2112 ⫹ 6 4
3
2
315
Figure 4.2
15
⫺4
4
⫺10
original function
substitute 1 for x
2 ⫽ 2a
substitute 2 for p(1), simplify
1⫽a
result
Consistent with the graph shown, if the point (1, 4) were specified instead, a like calculation would show a ⫽ 2.
EXAMPLE 8
䊳
Finding Zeroes Using the F actor Theorem
Given that 2 is a zero of p1x2 ⫽ x4 ⫹ x3 ⫺ 10x2 ⫺ 4x ⫹ 24, use the factor theorem
to help find all other zeroes.
Solution
䊳
Using synthetic division gives:
use 2 as a “divisor”
2
1
↓
1
1
2
3
⫺10
6
⫺4
⫺4
⫺8
⫺12
Since the remainder is zero, 1x ⫺ 22 is a factor and p can be written:
24
⫺24
0
x4 ⫹ x3 ⫺ 10x2 ⫺ 4x ⫹ 24 ⫽ 1x ⫺ 22 1x3 ⫹ 3x2 ⫺ 4x ⫺ 122
WORTHY OF NOTE
In Appendix A.4 we noted a
third degree polynomial
ax3 ⫹ bx2 ⫹ cx ⫹ d is factorable if
ad ⫽ bc. In Example 8,
11⫺122 ⫽ 31⫺42 and the polynomial
is factorable.
C. You’ve just seen how
we can use the factor theorem
to factor and build polynomials
Note the quotient polynomial can be factored by grouping to find the remaining
factors of p.
x4 ⫹ x3 ⫺ 10x2 ⫺ 4x ⫹ 24 ⫽
⫽
⫽
⫽
⫽
1x ⫺ 22 1x3 ⫹ 3x2 ⫺ 4x ⫺ 122
1x ⫺ 22 3 x2 1x ⫹ 32 ⫺ 41x ⫹ 32 4
1x ⫺ 22 3 1x ⫹ 32 1x2 ⫺ 42 4
1x ⫺ 22 1x ⫹ 32 1x ⫹ 22 1x ⫺ 22
1x ⫹ 32 1x ⫹ 22 1x ⫺ 22 2
group terms (in color)
remove common
factors from each group
factor common binomial
factor difference of squares
completely factored form
The final result shows 1x ⫺ 22 is actually a repeated factor, and the remaining
zeroes of p are ⫺3 and ⫺2.
Now try Exercises 65 through 78
䊳
D. Applications
While the factor and remainder theorems are valuable tools for factoring higher degree
polynomials, each has applications that extend beyond this use.
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EXAMPLE 9
䊳
Using the Remainder Theorem to Solve a Discharge Rate Application
The discharge rate of a river is a measure of
the river’s water flow as it empties into a
lake, sea, or ocean. The rate depends on
many factors, but is primarily influenced
by the precipitation in the surrounding area
and is often seasonal. Suppose the discharge
rate of the Shimote River was modeled by
D1m2 m4 22m3 147m2 317m 150, where D(m) represents the
discharge rate in thousands of cubic meters of
water per second in month m 1m 1 S Jan2.
a. What was the discharge rate in June
(summer heat)?
b. Is the discharge rate higher in February
(winter runoff) or October (fall rains)?
Solution
䊳
a. To find the discharge rate in June, we
evaluate D at m 6.
Using the remainder theorem gives
6
1
↓
1
22
6
16
147
96
51
317
306
11
150
66
216
In June, the discharge rate is 216,000 m3/sec.
b. For the discharge rates in February 1m 22 and October 1m 102, we have
2 1 22 147
317 150
40 214 206
↓ 2
103 356
1 20 107
D. You’ve just seen how
we can solve applications
using the remainder theorem
10 1
22 147
317 150
120 270 470
T 10
12 27
47 620
1
The discharge rate during the fall rains in October is much higher: 620 7 356.
Now try Exercises 81 through 84
䊳
4.1 EXERCISES
䊳
CONCEPTS AND VOCABULAR Y
Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.
1. For
division, we use the
the divisor to begin.
of
2. If the
is zero after division, then the
is a factor of the dividend.
3. If polynomial P(x) is divided by a linear divisor of
the form x c, the remainder is identical to
. This is a statement of the
theorem.
4. If P1c2 0, then
must be a factor of
P(x). Conversely, if
is a factor of P(x),
then P1c2 0. These are statements from the
theorem.
5. Discuss/Explain how to write the quotient and
remainder using the last line from a synthetic
division.
6. Discuss/Explain why (a, b) is a point on the graph
of P, given b was the remainder after P was divided
by x a using synthetic division.
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䊳
Section 4.1 Synthetic Division; the Remainder and Factor Theorems
317
DEVELOPING YOUR SKILLS
Divide using long division. Write the result as
dividend ⴝ (divisor)(quotient) ⴙ remainder.
Compute each indicated quotient. Write answers in the
remainder
form dividend
divisor ⴝ quotient ⴙ divisor .
7.
x3 5x2 4x 23
x2
31.
2x3 7x2 x 26
x4 3x3 2x2 x 5
32.
x2 3
x2 2
8.
x3 5x2 17x 26
x7
33.
x4 5x2 4x 7
x2 1
9. 12x3 5x2 4x 172 1x 32
35. P1x2 x3 6x2 5x 12
a. P122
b. P152
11. 1x 8x 11x 202 1x 52
2
12. 1x3 5x2 22x 162 1x 22
Divide using synthetic division. Write answers in
remainder
two ways: (a) dividend
divisor ⴝ quotient ⴙ divisor , and
(b) dividend ⫽ (divisor)(quotient) ⫹ remainder. For
Exercises 13–18, check answers using multiplication.
2x2 5x 3
13.
x3
3x2 13x 10
14.
x5
15. 1x3 3x2 14x 82 1x 22
16. 1x3 6x2 24x 172 1x 12
x 5x 4x 23
x 12x 34x 9
18.
x2
x7
3
17.
2
3
2
19. 12x 5x 11x 172 1x 42
3
2
20. 13x3 x2 7x 272 1x 12
Divide using synthetic division. Note that some terms of
a polynomial may be “missing.” Write answers as
dividend ⴝ (divisor)(quotient) ⴙ remainder.
21. 1x3 5x2 72 1x 12
22. 1x3 3x2 372 1x 52
23. 1x3 13x 122 1x 42
24. 1x3 7x 62 1x 32
25.
3x3 8x 12
x1
27. 1n3 272 1n 32
26.
2x3 7x 81
x3
28. 1m3 82 1m 22
29. 1x4 3x3 16x 82 1x 22
30. 1x4 3x2 29x 212 1x 32
x4 2x3 8x 16
x2 5
Use the remainder theorem to evaluate P(x) as given.
10. 13x3 14x2 2x 372 1x 42
3
34.
36. P1x2 x3 4x2 8x 15
a. P122
b. P132
37. P1x2 2x3 x2 19x 4
a. P132
b. P122
38. P1x2 3x3 8x2 14x 9
a. P122
b. P142
39. P1x2 x4 4x2 x 1
a. P122
b. P122
40. P1x2 x4 3x3 2x 4
a. P122
b. P122
41. P1x2 2x3 7x 33
a. P122
b. P132
42. P1x2 2x3 9x2 11
a. P122
b. P112
43. P1x2 2x3 3x2 9x 10
a. P1 32 2
b. P152 2
44. P1x2 3x3 11x2 2x 16
a. P1 13 2
b. P183 2
Use the factor theorem to determine if the factors given
are factors of f(x).
45. f 1x2 x3 3x2 13x 15
a. 1x 32
b. 1x 52
46. f 1x2 x3 2x2 11x 12
a. 1x 42
b. 1x 32
47. f 1x2 x3 6x2 3x 10
a. 1x 22
b. 1x 52
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48. f 1x2 x3 2x2 5x 6
a. 1x 22
b. 1x 42
In Exercises 65 through 70, a known zero of the
polynomial is given. Use the factor theorem to write the
polynomial in completely factored form.
49. f 1x2 2x3 x2 12x 9
a. 1x 32
b. 12x 32
65. P1x2 x3 5x2 2x 24; x 2
66. Q1x2 x3 7x2 7x 15; x 3
50. f 1x2 3x3 19x2 30x 8
a. 13x 12
b. 1x 42
67. p1x2 x4 2x3 12x2 18x 27; x 3
68. q1x2 x4 4x3 6x2 4x 5; x 1
Use the factor theorem to show the given value is a zero
of P(x).
52. P1x2 x3 3x2 16x 12; x 6
53. P1x2 x 7x 6; x 2
3
54. P1x2 x3 13x 12; x 4
55. P1x2 9x3 18x2 4x 8; x 2
3
56. P1x2 5x3 13x2 9x 9; x 69. f 1x2 2x3 11x2 x 30; x 32
70. g1x2 3x3 2x2 75x 50; x 23
51. P1x2 x3 2x2 5x 6; x 3
If p(x) is a polynomial with rational coefficients and a
leading coefficient of a ⴝ 1, the rational zeroes of p (if
they exist) must be factors of the constant term. Use this
property of polynomials with the factor and remainder
theorems to factor each polynomial completely.
71. p1x2 x3 3x2 9x 27
3
5
A polynomial P with integer coefficients has the zeroes
and degree indicated. Use the factor theorem to write
the function in factored form and standard form.
57. 2, 3, 5; degree 3
58. 1, 4, 2; degree 3
59. 2, 23, 23;
degree 3
60. 25, 25, 4;
degree 3
61. 5, 2 23, 2 23;
degree 3
62. 4, 3 22, 322;
degree 3
63. 1, 2, 210, 210;
degree 4
64. 27, 27, 3, 1;
degree 4
䊳
4–12
CHAPTER 4 Polynomial and Rational Functions
72. p1x2 x3 4x2 16x 64
73. p1x2 x3 6x2 12x 8
74. p1x2 x3 15x2 75x 125
75. p1x2 1x2 6x 92 1x2 92
76. p1x2 1x2 12 1x2 2x 12
77. p1x2 1x3 4x2 9x 362 1x2 x 122
78. p1x2 1x3 3x2 3x 12 1x2 3x 22
WORKING WITH FORMULAS
Volume of an open box: V(x) ⴝ 4x3 ⴚ 84x2 ⴙ 432x
An open box is constructed by cutting square corners from a 24 in. by 18 in. sheet of cardboard
and folding up the sides. Its volume is given by the formula shown, where x represents the length
of the square cuts.
79. Given a volume of 640 in3, use synthetic division
and the remainder theorem to determine if the
squares were 2-, 3-, 4-, or 5-in. squares and state
the dimensions of the box. (Hint: Write as a
function V(x) and use synthetic division.)
80. Given the volume is 357.5 in3, use synthetic
division and the remainder theorem to determine if
the squares were 5.5-, 6.5-, or 7.5-in. squares and
state the dimensions of the box. (Hint: Write as a
function V(x) and use synthetic division.)
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䊳
Section 4.1 Synthetic Division; the Remainder and Factor Theorems
APPLICATIONS
81. Tourist population:
During the 12 weeks
of summer, the
population of
tourists at a popular
beach resort is
modeled by the
polynomial
P1w2 0.1w4 2w3 14w2 52w 5,
where P1w2 is the tourist population (in 1000s)
during week w. Use the remainder theorem to help
answer the following questions.
a. Were there more tourists at the resort in week 5
1w 52 or week 10? How many more tourists?
b. Were more tourists at the resort one week after
opening 1w 12 or one week before closing
1w 112. How many more tourists?
c. The tourist population peaked (reached its
highest) between weeks 7 and 10. Use the
remainder theorem to determine the peak
week.
82. Debt load: Due to a fluctuation in tax revenues, a
county government is projecting a deficit for the next
12 months, followed by a quick recovery and the
repayment of all debt near the end of this period. The
projected debt can be modeled by the polynomial
D1m2 0.1m4 2m3 15m2 64m 3, where
D1m2 represents the amount of debt (in millions of
dollars) in month m. Use the remainder theorem to
help answer the following questions.
a. Was the debt higher in month 5 1m 52 or
month 10 of this period? How much higher?
b. Was the debt higher in the first month of this
period (one month into the deficit) or after the
eleventh month (one month before the
expected recovery)? How much higher?
䊳
319
c. The total debt reached its maximum between
months 7 and 10. Use the remainder theorem
to determine which month.
83. Volume of water: The volume of water in a
rectangular, inground, swimming pool is given by
V1x2 x3 11x2 24x, where V(x) is the volume
in cubic feet when the water is x ft high. (a) Use
the remainder theorem to find the volume when
x 3 ft. (b) If the volume is 100 ft3 of water, what
is the height x? (c) If the maximum capacity of the
pool is 1000 ft3, what is the maximum depth (to the
nearest integer)?
84. Amusement park attendance: Attendance at an
amusement park depends on the weather. After
opening in spring, attendance rises quickly, slows
during the summer, soars in the fall, then quickly
falls with the approach of winter when the park
closes. The model for attendance is given by
A1m2 14 m4 6m3 52m2 196m 260,
where A(m) represents the number of people
attending in month m (in thousands). (a) Did more
people go to the park in April 1m 42 or June
1m 62? (b) In what month did maximum
attendance occur? (c) When did the park close?
In these applications, synthetic division is applied in the
usual way, treating k as an unknown constant.
85. Find a value of k that will make x 2 a factor of
f 1x2 x3 3x2 5x k.
86. Find a value of k that will make x 3 a factor of
g1x2 x3 2x2 7x k.
87. For what value(s) of k will x 2 be a factor of
p1x2 x3 3x2 kx 10?
88. For what value(s) of k will x 5 be a factor of
q1x2 x3 6x2 kx 50?
EXTENDING THE CONCEPT
89. To investigate whether the remainder and factor
theorems can be applied when the coefficients or
zeroes of a polynomial are complex, try using the
factor theorem to find a polynomial with degree 3,
whose zeroes are x 2i, x 2i, and x 3.
Then see if the result can be verified using the
remainder theorem and these zeroes. What does the
result suggest? Also see Exercise 92.
90. Since we use a base-10 number system, numbers
like 1196 can be written in polynomial form as
p1x2 1x3 1x2 9x 6, where x 10. Divide
p(x) by x 3 using synthetic division and write
3
2
9x 6
quotient your answer as x xx 3
remainder
divisor . For x 10, what is the value of
quotient remainder
divisor ? What is the result of dividing
1196 by 10 3 13? What can you conclude?
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91. The sum of the first n perfect cubes is given by the
formula S 14 1n4 2n3 n2 2. Use the remainder
theorem on S to find the sum of (a) the first three
perfect cubes (divide by n 32 and (b) the first
five perfect cubes (divide by n 52. Check results
by adding the perfect cubes manually. To avoid
working with fractions you can initially ignore the
1
4
3
2
4 (use n 2n n 0n 02, as long as you
divide the remainder by 4.
92. Though not a direct focus of this course, the
remainder and factor theorems, as well as synthetic
䊳
division, can also be applied using complex
numbers. Use the remainder theorem to show the
value given is a zero of P(x).
a. P1x2 x3 4x2 9x 36; x 3i
b. P1x2 x4 x3 2x2 4x 8; x 2i
c. P1x2 x3 x2 3x 5; x 1 2i
d. P1x2 x3 2x2 16x 32; x 4i
e. P1x2 x4 x3 5x2 x 6; x i
f. P1x2 x3 x2 8x 10; x 1 3i
MAINTAINING YOUR SKILLS
93. (1.5) John and Rick are out orienteering. Rick finds
the last marker first and is heading for the finish
line, 1275 yd away. John is just seconds behind,
and after locating the last marker tries to overtake
Rick, who by now has a 250-yd lead. If Rick runs
at 4 yd/sec and John runs at 5 yd/sec, will John
catch Rick before they reach the finish line?
94. (Appendix A.3) Solve for w:
213w2 52 3 7w w2 7
4.2
In Section 4.2 you will see
how we can:
A. Apply the fundamental
C.
D.
E.
96. (3.4) Given f 1x2 x2 4x, use the average rate of
change formula to find ¢y
¢x in the interval
x 僆 31.0, 1.1 4.
The Zeroes of Polynomial Functions
LEARNING OBJECTIVES
B.
95. (1.4) The profit of a small business increased
linearly from $5000 in 2005 to $12,000 in 2010.
Find a linear function G(t) modeling the growth of
the company’s profit (let t 0 correspond to
2005).
theorem of algebra and
the linear factorization
theorem
Use the intermediate
value theorem to identify
intervals containing a
polynomial zero
Find rational zeroes of a
real polynomial function
using the rational zeroes
theorem
Obtain more information
on the zeroes of real
polynomials using
Descartes’ rule of signs
and the upper/lower
bounds theorem
Solve applications of
polynomial functions
This section represents one of the highlights in the college algebra curriculum, because
it offers a look at what many call the big picture. The ideas presented are the result of
a cumulative knowledge base developed over a long period of time, and give a fairly
comprehensive view of the study of polynomial functions.
A. The Fundamental Theorem of Algebra
From Section 3.1, we know the set of real numbers is a subset of the complex numbers.
Because complex numbers are the “larger” set (containing all other number sets), properties and theorems about complex numbers are more powerful and far reaching than
theorems about real numbers. In the same way, real polynomials are a subset of the
complex polynomials, and the same principle applies.
Complex Polynomial Functions
A complex polynomial of degree n has the form
P1x2 anxn an1xn1 p a1x1 a0,
where an, an1, p , a1, a0 are complex numbers and an 0.
Notice that real polynomials have the same form, but here an, an1, p , a1, a0 represent complex numbers. In 1799, Carl Friedrich Gauss (1777–1855) proved that all
polynomial functions have zeroes, and that the number of zeroes is equal to the degree
of the polynomial. The proof of this statement is based on a theorem that is the bedrock
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321
for a complete study of polynomial functions, and has come to be known as the fundamental theorem of algebra.
WORTHY OF NOTE
Quadratic functions also
belong to the larger family of
complex polynomial functions.
Since quadratics have a known
number of terms, it is common to
write the general form using the
early letters of the alphabet:
P1x2 ⫽ ax2 ⫹ bx ⫹ c. For higher
degree polynomials, the number of
terms is unknown or unspecified,
and the general form is written
using subscripts on a single letter.
The Fundamental Theorem of Algebra
Every complex polynomial of degree n ⱖ 1 has at least one complex zero.
Although the statement may seem trivial, it allows us to draw two important
conclusions. The first is that our search for a solution will not be fruitless or wasted—
zeroes for all polynomial equations exist. Second, the fundamental theorem combined
with the factor theorem enables us to state the linear factorization theorem.
The Linear Factorization Theorem
If p(x) is a polynomial function of degree n ⱖ 1, then p has exactly n linear factors
and can be written in the form,
p1x2 ⫽ a1x ⫺ c1 2 1x ⫺ c2 2 # p # 1x ⫺ cn 2
where a ⫽ 0 and c1, c2, p , cn are (not necessarily distinct) complex numbers.
In other words, every complex polynomial of degree n can be rewritten as the
product of a nonzero constant and exactly n linear factors (for a proof of this theorem,
see Appendix B).
EXAMPLE 1
䊳
Writing a Polynomial as a Product of Linear F actors
Rewrite P1x2 ⫽ x4 ⫺ 8x2 ⫺ 9 as a product of linear factors, and find its zeroes.
Solution
䊳
From its given form, we know a ⫽ 1. Since P has degree 4, the factored form must
be P1x2 ⫽ 1x ⫺ c1 2 1x ⫺ c2 2 1x ⫺ c3 2 1x ⫺ c4 2 . Noting that P is in quadratic form,
we substitute u for x2 and u2 for x4 and attempt to factor:
x4 ⫺ 8x2 ⫺ 9 S u2 ⫺ 8u ⫺ 9
⫽ 1u ⫺ 92 1u ⫹ 12
⫽ 1x2 ⫺ 92 1x2 ⫹ 12
WORTHY OF NOTE
While polynomials with complex
coefficients are not the focus of this
course, interested students can
investigate the wider application of
these theorems by completing
Exercise 115.
substitute u for x 2; u 2 for x 4
factor in terms of u
rewrite in terms of x (substitute x 2 for u )
We know x2 ⫺ 9 will factor since it is a difference of squares. From our work with
complex numbers (Section 3.1), we know 1a ⫹ bi2 1a ⫺ bi2 ⫽ a2 ⫹ b2, and the
factored form of x2 ⫹ 1 must be 1x ⫹ i2 1x ⫺ i2 . The completely factored form is
P1x2 ⫽ 1x ⫹ 32 1x ⫺ 32 1x ⫹ i2 1x ⫺ i2, and
the zeroes of P are ⫺3, 3, ⫺i, and i.
Now try Exercises 7 through 10
EXAMPLE 2
䊳
䊳
Writing a Polynomial as a Product of Linear F actors
Rewrite P1x2 ⫽ x3 ⫹ 2x2 ⫺ 4x ⫺ 8 as a product of linear factors and find its zeroes.
Solution
䊳
We observe that a ⫽ 1 and P has degree 3, so the factored form must be
P1x2 ⫽ 1x ⫺ c1 2 1x ⫺ c2 2 1x ⫺ c3 2 . Noting that ad ⫽ bc (Appendix A.4, page A-38),
we start with factoring by grouping.
P1x2 ⫽ x3 ⫹ 2x2 ⫺ 4x ⫺ 8
⫽ x2 1x ⫹ 22 ⫺ 41x ⫹ 22
⫽ 1x ⫹ 22 1x2 ⫺ 42
⫽ 1x ⫹ 22 1x ⫹ 22 1x ⫺ 22
group terms (in color)
remove common factors (note sign change)
factor common binomial
factor difference of squares
The zeroes of P are ⫺2, ⫺2, and 2.
Now try Exercises 11 through 14
䊳
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Note the polynomial in Example 2 has three zeroes, but the zero 2 was repeated
two times. In this case we say 2 is a zero of multiplicity two, and a zero of even multiplicity. It is also possible for a zero to be repeated three or more times, with those repeated an odd number of times called zeroes of odd multiplicity [the factor
1x 22 1x 22 1 also gives a zero of odd multiplicity]. In general, repeated factors
are written in exponential form and we have
Zeroes of Multiplicity
If p is a polynomial function with degree n 1, and (x c) occurs as a factor of p
exactly m times, then c is a zero of multiplicity m.
EXAMPLE 3
Solution
䊳
Identifying the Multiplicity of a Zero
Factor the given function completely, writing repeated factors in exponential form.
Then state the multiplicity of each zero: P1x2 1x2 8x 162 1x2 x 2021x 52
䊳
P1x2 1x2 8x 162 1x2 x 202 1x 52
1x 42 1x 42 1x 52 1x 42 1x 52
1x 42 3 1x 52 2
given polynomial
trinomial factoring
exponential form
For function P, 4 is a zero of multiplicity 3 (odd multiplicity), and 5 is a zero of
multiplicity 2 (even multiplicity).
Now try Exercises 15 through 18
䊳
These examples help illustrate three important consequences of the linear
factorization theorem. From Example 1, if the coefficients of P are real, the polynomial
can be factored into linear and quadratic factors using real numbers only
3 1x 32 1x 32 1x2 12 4 , where the quadratic factors have no real zeroes. Quadratic
factors of this type are said to be irreducible.
Corollary I: Irreducible Quadratic Factors
If p is a polynomial with real coefficients, p can be factored into a product of linear
factors (which are not necessarily distinct) and irreducible quadratic factors having
real coefficients.
Closely related to this corollary and our previous study of quadratic functions,
complex zeroes of the irreducible factors must occur in conjugate pairs.
Corollary II: Complex Conjugates
If p is a polynomial with real coefficients, complex zeroes must occur in conjugate
pairs. If a bi, b 0 is a zero, then a bi will also be a zero.
Finally, the polynomial in Example 1 has degree 4 with 4 zeroes (two real, two
complex), and the polynomial in Example 2 has degree 3 with 3 zeroes (three real, one
of these is a repeated root). While not shown explicitly, the polynomial in Example 3
has degree 5, and there were 5 zeroes (one repeated twice, one repeated three times).
This suggests our final corollary.
Corollary III: Number of Zeroes
If p is a polynomial function with degree n 1, then p has exactly n zeroes
(real or complex), where zeroes of multiplicity m are counted m times.
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323
These corollaries help us gain valuable information about a polynomial, when
only partial information is given or known.
EXAMPLE 4
䊳
Constructing a Polynomial from Its Zeroes
A polynomial P of degree 3 with real coefficients has zeroes of 1 and 2 i13.
Find the polynomial (assume a 1).
Solution
䊳
WORTHY OF NOTE
Using the factor theorem, two of the factors are (x 1) and x 12 i 132 .
From Corollary II, 2 i 13 must also be a zero and x 12 i132 is also a
factor of P. This gives
P1x2 1x 12 3 x 12 i 132 4 3 x 12 i 132 4
1x 12 3 1x 22 i 134 3 1x 22 i 134
1x 12 3 1x2 4x 42 3 4
1x 12 1x2 4x 72
x3 3x2 3x 7
When reconstructing a polynomial
P having complex zeroes, it is often
more efficient to determine the
irreducible quadratic factors of P
separately, as shown here. For the
zeroes 2 i 13 we have
x 2 i13
x 2 i13
1x 22 2 1i132 2
2
x 4x 4 3
x2 4x 7 0.
The quadratic factor is
1x2 4x 72 .
associative property
1a bi 21a bi 2 a 2 b 2
simplify
result
The polynomial is P1x2 x 3x 3x 7,
which can be verified using the remainder
theorem and any of the original zeroes. A
calculator check using the zero x 2 i 23 is
shown here.
3
2
Now try Exercises 19 through 22
EXAMPLE 5
䊳
䊳
Building a Polynomial from Its Zeroes
Find a fourth degree polynomial P with real coefficients, if 3 is the only real zero
and 2i is also a zero of P.
Solution
䊳
Since complex zeroes must occur in conjugate pairs, 2i is also a zero, but this
accounts for only three zeroes. Since P has degree 4, 3 must be a repeated zero,
and the factors of P are 1x 32 1x 32 1x 2i2 1x 2i2 .
P1x2 1x 32 1x 32 1x 2i2 1x 2i2
1x2 6x 92 1x2 42
x4 6x3 13x2 24x 36
A. You’ve just seen how
we can apply the fundamental
theorem of algebra and the
linear factorization theorem
factored form
multiply binomials, 1a bi 21a bi 2 a 2 b 2
result
The polynomial is P1x2 x4 6x3 13x2 24x 36, which can be verified
using a calculator or the remainder theorem and any of the original zeroes.
Now try Exercises 23 through 28
䊳
B. Real Polynomials and the Intermediate Value Theorem
The fundamental theorem of algebra is called an existence theorem, as it affirms
the existence of the zeroes but does not tell us where or how to find them. Because
polynomial graphs are continuous (there are no holes or breaks in the graph), the
intermediate value theorem (IVT) can be used for this purpose.
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CHAPTER 4 Polynomial and Rational Functions
The Intermediate Value Theorem
Given P is a polynomial with real coefficients, if P(a) and P(b) have opposite signs,
there is at least one number c between a and b such that P1c2 0 (see Figure 4.3).
Figure 4.3
y
P(b) 0
(b, P(b))
WORTHY OF NOTE
You might recall a similar idea was
used in Section 2.1, where we
noted the graph of P(x) crosses the
x-axis at the zeroes determined by
linear factors, with a corresponding
change of sign in the function
values.
Intermediate
value
a
P(c) 0
c
x
b
P(a) 0
(a, P(a))
EXAMPLE 6
Solution
䊳
Finding Zeroes Using the Inter mediate Value Theorem
Use the intermediate value theorem to show P1x2 x3 9x 6 has at least one
zero in the interval given:
a. 34, 34
b. [0, 1]
c. [4, 3]
䊳
a. Begin by evaluating P at x 4 and x 3.
P142 142 3 9142 6
64 36 6
22
P132 132 3 9132 6
27 27 6
6
Since P(4) 0 and P132 7 0, there must be at least one number c1
between 4 and 3 where P1c1 2 0. The graph must cross the x-axis at least
once in this interval.
b. Evaluate P at x 0 and x 1.
P102 102 3 9102 6
006
6
P112 112 3 9112 6
196
2
Since P102 7 0 and P112 6 0, there must be at least one number c2 between 0
and 1 where P1c2 2 0.
c. In part (a) we found that P142 6 0. For P(3) we have
P132 132 3 9132 6
27 27 6
6
Since P142 6 0 and P132 7 0, there
must be at least one number c3 between
4 and 3 where P1c3 2 0. In fact, from
parts (a) and (b) we know of two that exist,
and the graph of P shown in Figure 4.4
reveals there are actually three real roots
in this interval. This illustrates why the
intermediate value theorem uses the
phrase, “there is at least one number c.”
Figure 4.4
20
5
5
10
Now try Exercises 29 and 30
䊳
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Section 4.2 The Zeroes of Polynomial Functions
Figure 4.4A
To help illustrate the Intermediate Value Theorem, many graphing calculators
offer a useful feature called split screen viewing, that enables us to view a table of
values and the graph of a function at the same time. To illustrate, enter the function
y ⫽ x3 ⫺ 9x ⫹ 6 (from Example 6) as Y1 on the Y= screen, then set the viewing
window as shown in Figure 4.4. Set your table in AUTO mode with ¢Tbl ⫽ 1, then
press the MODE key (see Figure 4.4A) and notice the second-to-last entry on this screen
reads: Full for full screen viewing, Horiz for splitting the screen horizontally with the
graph above a reduced home screen, and G-T, which represents Graph-Table and
splits the screen vertically. In the G-T mode, the graph appears on the left and the
table of values on the right. Navigate the cursor to the G-T mode and press . Pressing the GRAPH key at this point should give you a screen similar to Figure 4.5. Scrolling
downward shows the function also changes sign between x ⫽ 2 and x ⫽ 3. For more
on this idea, see Exercises 31 and 32.
As a final note, while the intermediate value theorem is a powerful yet simple
tool, it must be used with care. For example, given p1x2 ⫽ ⫺x4 ⫹ 10x2 ⫺ 5, p1⫺12 7 0
and p112 7 0, seeming to indicate that no zeroes exist in the interval (⫺1, 1). Actually,
there are two zeroes, as seen in Figure 4.6.
ENTER
Figure 4.6
Figure 4.5
25
B. You’ve just seen how
we can use the intermediate
value theorem to identify
intervals containing a
polynomial zero
⫺5
5
⫺10
C. The Rational Zeroes Theorem
The fundamental theorem of algebra tells us that zeroes of a polynomial function
exist. The intermediate value theorem tells us how to locate intervals that contain
zeroes. Our next theorem gives us the information we need to actually find certain
zeroes of a polynomial. Recall that if c is a zero of P, then P1c2 ⫽ 0, and when P(x)
is divided by x ⫺ c using synthetic division, the remainder is zero (from the remainder and factor theorems).
To find divisors that give a remainder of zero, we make the following observations. To solve 3x2 ⫺ 11x ⫺ 20 ⫽ 0 by factoring, a beginner might write out all possible binomial pairs where the First term in the F-O-I-L process multiplies to 3x2 and
the Last term multiplies to 20. The six possibilities are shown here:
(3x
1)(x
20)
(3x
20)(x
1)
(3x
2)(x
10)
(3x
4)(x
5)
(3x
5)(x
4)
(3x
10)(x
2)
If 3x ⫺ 11x ⫺ 20 is factorable using integers, the factors must be somewhere
in this list. Also, the first coefficient in each binomial must be a factor of the leading
coefficient, and the second coefficient must be a factor of the constant term. This
means that regardless of which factored form is correct, the solution will be a rational
number whose numerator comes from the factors of 20, and whose denominator
comes from the factors of 3. The correct factored form is shown here, along with
the solution:
2
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3x2 11x 20 0
13x 42 1x 52 0
3x 4 0
4 d from the factors of 20
x
3 d from the factors of 3
x50
5 d from the factors of 20
x
1 d from the factors of 3
This same principle also applies to polynomials of higher degree, and these observations suggest the following theorem.
The Rational Zeroes Theorem
p
Given polynomial P with integer coefficients, and a rational number in lowest terms,
q
p
the rational zeroes of P (if they exist) must be of the form , where p is a factor of the
q
constant term, and q is a factor of the leading coefficient.
Note that if the leading coefficient is 1, the possible rational zeroes are limited to
p
factors of the constant term: 1 p. If the leading coefficient is not “1” and the constant
term has a large number of factors, the set of possible rational zeroes becomes rather
large. To list these possibilities, it helps to begin with all factor pairs of the constant a0,
then divide each of these by the factors of an as shown in Example 7.
EXAMPLE 7
䊳
Identifying the Possible Rational Zeroes of a P olynomial
List all possible rational zeroes for the function P1x2 3x4 14x3 x2 42x 24.
Solution
䊳
All rational zeroes must be of the form pq , where p is a factor of a0 24 and q is
a factor of an 3. The factor pairs of 24 are: 1, 24, 2, 12, 3, 8, 4
and 6. Dividing each by 1 and 3 (the factor pairs of 3), we note division by
1 will not change any of the previous values, while division by 3 gives
13, 23, 83, 43 as additional possibilities. Any rational zeroes must be from the set
51, 24, 2, 12, 3, 8, 4, 6, 13, 23, 83,43 6 .
Now try Exercises 33 through 40
䊳
The actual zeroes of the function in Example 7 are x 13, x 13, x 23,
and x 4 and the graph of P1x2 3x4 14x3 x2 42x 24 is shown in
Figure 4.7. Although the rational zeroes are inFigure 4.7
deed in the set noted, it’s apparent we need a way
60
to narrow down the number of possibilities (we
don’t want to try all 24 possible zeroes). If we’re
able to find even one factor easily, we can rewrite
the polynomial using this factor and the quotient 6
3
polynomial, with the hope of factoring further
using trinomial factoring or factoring by grouping. Many times testing to see if 1 or 1 are
zeroes will help.
60
Tests to Determine If 1 or ⫺1 is a Zero of P
For any polynomial P with real coefficients,
1. If the sum of all coefficients is zero, then 1 is a root and 1x 12 is a factor.
2. After changing the sign of all terms with odd degree, if the sum of the
coefficients is zero, then 1 is a root and 1x 12 is a factor.
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Section 4.2 The Zeroes of Polynomial Functions
EXAMPLE 8
327
Finding the Rational Zeroes of a P olynomial
䊳
Find all rational zeroes of P1x2 3x4 x3 8x2 2x 4, and use them to write
the function in completely factored form. Then use the factored form to name all
zeroes of P.
Solution
䊳
Instead of listing all possibilities using the rational zeroes theorem, we first test for
1 and 1, then see if we’re able to complete the factorization using other means.
The sum of the coefficients is: 3 1 8 2 4 0, which means 1 is a zero
and x 1 is a factor. By changing the sign on terms of odd degree, we have
3x4 x3 8x2 2x 4 and 3 1 8 2 4 2, showing 1 is not a
zero. Using x 1 and the factor theorem, we have
WORTHY OF NOTE
In the second to last line of
Example 8, we factored x2 2 as
(x 1221x 12). As discussed in
Appendix A.6, this is an application
of factoring the difference of two
squares: a2 b2 1a b21a b2 .
By mentally rewriting x2 2 as
x2 1 122 2, we obtain the result
shown. Also see Exercise 113.
use 1 as a “divisor”
1
3
2
1 3
3
8
2
6
2
6
4
4
4
0
and we write P as P1x2 1x 12 13x3 2x2 6x 42 . Noting the quotient
polynomial can be factored by grouping (ad bc), we need not continue with
synthetic division or the factor theorem.
8
3
3
4
P1x2 1x 12 13x3 2x2 6x 42
1x 12 3 x2 13x 22 213x 22 4
1x 12 13x 22 1x2 22
1x 12 13x 22 1x 122 1x 122
The zeroes of P are 1,
2
3 ,
group terms
factor common terms
factor common binomial
completely factored form
and 12. The graph of P is shown in the figure.
Now try Exercises 41 through 62
䊳
In cases where the quotient polynomial is not easily factored, we continue with
synthetic division and other possible zeroes, until the remaining zeroes can be
determined.
EXAMPLE 9
䊳
Finding the Zeroes of a P olynomial
Find all zeroes of P1x2 x5 3x4 3x3 5x2 12.
Solution
䊳
Using the rational zeroes theorem, the possibilities are: 51, 12, 2, 6, 3, 46.
The test for 1 shows 1 is not a zero. After changing the signs of all terms with odd
degree, we have 1 3 3 5 12 0, and find 1 is a zero. Using 1
with the factor theorem, we continue our search for additional factors. Noting that
P is missing a linear term, we include a placeholder zero:
use 1 as a “divisor”
1
1
1
3
1
4
3
4
7
5
7
12
0
12
12
12
12
0
coefficients of P
coefficients of q1(x )
Here the quotient polynomial q1 1x2 x4 4x3 7x2 12x 12 is not easily
factored, so we next try 2, using the quotient polynomial:
use 2 as a “divisor” on q1(x )
2
1
1
4
2
2
7
4
3
12
6
6
12
12
0
coefficients of q1(x )
coefficients of q2(x )
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If you miss the fact that q2(x) is actually factorable (ad bc), the process would
continue using 2 and the current quotient.
use 2 as a “divisor”
2
1
1
2
2
4
3
8
11
6
22
28
coefficients of q2 (x )
2 is not a zero
We find 2 is not a zero, and in fact, trying all other possibilities will show that
none of them are zeroes. As there must be five zeroes, we are reminded of three
things:
1. This process can only find rational zeros (the remaining zeroes may be
irrational or complex),
2. This process cannot find irreducible quadratic factors (unless they appear as the
quotient polynomial), and
3. Some of the zeroes may have multiplicities greater than 1!
Testing the zero 2 for a second time using q2(x) gives
use 2 as a “divisor”
2
1
1
2
2
0
and we see that 2 is actually a zero of
multiplicity two, and the final quotient is
the irreducible quadratic factor x2 3.
Using this information produces the factored
form P1x2 1x 12 1x 22 2 1x2 32 1x 12 1x 22 2 1x i 132 1x i 132 ,
and the zeroes of P are i1 3 , i 1 3, 1,
and 2 with multiplicity two. The graph of P
is shown in the figure.
3
0
3
6
6
0
coefficients of q2(x )
2 is a repeated zero
24
3
4
10
Now try Exercises 63 through 82
C. You’ve just seen how
we can find rational zeroes of
a real polynomial function
using the rational zeroes
theorem
In Example 9, note that since the leading
coefficient is 1, all possible rational zeroes
will be integers. This means our initial
search can easily be performed using the
TABLE feature of a graphing calculator.
But note that while Figure 4.8 indeed
shows that x 1 and x 2 are zeroes, it
cannot show that x 2 is a repeated zero
until we begin working with the quotient
polynomial. In the next section, we’ll learn
how to locate repeated zeroes from a polynomial’s graph.
䊳
Figure 4.8
D. Descartes’ Rule of Signs and Upper/Lower Bounds
Testing x 1 and x 1 is one way to reduce the number of possible rational zeroes,
but unless we’re very lucky, factoring the polynomial can still be a challenge.
Descartes’ rule of signs and the upper and lower bounds property offer additional
assistance.
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Section 4.2 The Zeroes of Polynomial Functions
Descartes’ Rule of Signs
Given the real polynomial equation P1x2 0,
1. The number of positive real zeroes is equal to the number of variations in
sign for P(x), or an even number less.
2. The number of negative real zeroes is equal to the number of variations in
sign for P1x2 , or an even number less.
EXAMPLE 10
䊳
Finding the Zeroes of a P olynomial
For P1x2 2x5 5x4 x3 x2 x 6,
a. Use the rational zeroes theorem to list all possible rational zeroes.
b. Apply Descartes’ rule to count the number of possible positive, negative, and
complex zeroes.
c. Use this information and the tools of this section to find all zeroes of P.
Solution
䊳
a. The factors of 2 are 51, 26 and the factors of 6 are 51, 6, 2, 36 . The
possible rational zeroes for P are 51, 6, 2, 3, 12, 32 6 .
b. For Descartes’ rule, we organize our work in a table. Since P has degree 5,
there must be a total of five zeroes. For this illustration, positive terms are in
blue and negative terms in red: P1x2 2x5 5x4 x3 x2 x 6. The
terms change sign a total of four times, meaning there are four, two, or zero
positive roots. For the negative roots, recall that P1x2 will change the sign of
all odd-degree terms, giving P1x2 2x5 5x4 x3 x2 x 6. This
time there is only one sign change (from negative to positive) showing there
will be exactly one negative root, a fact that is highlighted in the following
table. Since there must be 5 zeroes, the number of possible complex zeroes is:
none, two, or four, as shown.
possible
positive zeroes
known
negative zeroes
possibilities for
complex roots
total number
must be 5
4
1
0
5
2
1
2
5
0
1
4
5
c. Testing 1 and 1 shows x 1 is not a root, but x 1 is, and using 1 in
synthetic division gives:
use 1 as a “divisor”
1
WORTHY OF NOTE
As you recall from our study of
quadratics, it’s entirely possible for
a polynomial function to have no
real zeroes. Also, if the zeroes are
irrational, complex, or a combination
of these, they cannot be found
using the rational zeroes theorem.
For a look at ways to determine
these zeroes, see the Reinforcing
Basic Skills feature that follows
Section 4.3.
5
2
7
2
2
1
7
8
1
8
7
1
7
6
6
6
0
coefficients of P (x )
q1(x ) is not easily factored
Since there is only one negative root, we need only check the remaining
positive zeroes. The quotient q1(x) is not easily factored, so we continue with
synthetic division using the next larger positive root, x 2.
use 2 as a “divisor”
2
2
2
7
4
3
8
6
2
7
4
3
6
6
0
coefficients of q1(x )
q2(x )
The partially factored form is P1x2 1x 12 1x 22 12x3 3x2 2x 32 .
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10
Graphing P(x) at this point (see the figure)
verifies that 1 and 2 are zeroes, but also
indicates there is an additional zero between
1 and 2. If it’s a rational zero, it
4
3
3
must be x from our list of possible zeroes.
2
3
Checking using synthetic division and
2
5
q2(x) we have
3
2 3 2 3
2
T
3 0
3
2
0 2
0
3
3
Since the remainder is zero, P a b 0 (remainder theorem) and ax b is a
2
2
factor (factor theorem). Since the zero is a fraction, we’ll use the ideas discussed in
Section 4.1 to help write P(x) in completely factored form:
3
P1x2 1x 12 1x 22ax b12x2 22
2
3
1x 12 1x 22ax b122 1x2 12
2
1x 12 1x 22 12x 32 1x2 12
1x 12 1x 22 12x 32 1x i2 1x i2
partially factored form
factor out 2
3
multiply 2 ax b
2
completely factored form
The zeroes of P are 1, 2, 32, i and i, with two positive, one negative, and two
complex zeroes (row two of the table).
Now try Exercises 83 through 96
䊳
One final idea that helps reduce the number of possible zeroes is the upper and
lower bounds property. A number b is an upper bound on the positive zeroes of a
function if no positive zero is greater than b. In the same way, a number a is a lower
bound on the negative zeroes if no negative zero is less than a.
Upper and Lower Bounds Proper ty
Given P(x) is a polynomial with real coefficients.
1. If P(x) is divided by x b 1b 7 02 using synthetic division and all
coefficients in the quotient row are either positive or zero, then b is
an upper bound on the zeroes of P.
2. If P(x) is divided by x a 1a 6 02 using synthetic division and all
coefficients in the quotient row alternate in sign, then a is a lower
bound on the zeroes of P.
For both 1 and 2, zero coefficients can be either positive or negative as needed.
D. You’ve just seen how we
can obtain more information on
the zeroes of real polynomials
using Descartes’ rule of signs
and upper/lower bounds
theorem
While this test certainly helps narrow the possibilities, we gain the additional
benefit of knowing the property actually places boundaries on all real zeroes of the
polynomial, both rational and irrational. In part (c) of Example 10, the quotient row
of the first division alternates in sign, showing x 1 is both a zero and a lower
bound on the real zeroes of P. For more on the upper and lower bounds property, see
Exercise 111.
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Section 4.2 The Zeroes of Polynomial Functions
E. Applications of Polynomial Functions
Polynomial functions can be very accurate models of real-world phenomena, though
we often must restrict their domain, as illustrated in Example 11.
EXAMPLE 11
䊳
Using the Remainder Theorem to Solve an Oceanograph y Application
As part of an environmental study, scientists use radar to map the ocean floor from
the coastline to a distance 12 mi from shore. In this study, ocean trenches appear as
negative values and underwater mountains as positive values, as measured from the
surrounding ocean floor. The terrain due west of a particular island can be modeled
by h1x2 x4 25x3 200x2 560x 384, where h(x) represents the height in
feet, x mi from shore 10 6 x 122.
a. Use the remainder theorem to find the “height of the ocean floor” 10 mi out.
b. Use the tools developed in this section to find the number of times the ocean
floor has height h1x2 0 in this interval, given this occurs 12 mi out.
Solution
䊳
a. For part (a) we simply evaluate h(10) using the remainder theorem.
use 10 as a “divisor”
10
1
1
25
10
15
200
150
50
560
500
60
384
600
216
coefficients of h(x)
remainder is 216
Ten miles from shore, there is an ocean trench 216 ft deep.
b. For part (b), we're given 12 is zero, so we again use the remainder theorem and
work with the quotient polynomial.
use 12 as a “divisor”
12
1
25
12
13
3
2
1
560
528
32
200
156
44
384
384
0
coefficients of h(x)
q1(x)
The quotient is q1 1x2 x 13x 44x 32. Since a 1, we know the
remaining rational zeroes must be factors of 32: 51, 32, 2, 16, 4, 86.
Using x 1 gives
use 1 as a “divisor”
1
1
1
13
1
12
44
12
32
32
32
0
coefficients of q1(x)
q2(x)
The function can now be written as h1x2 1x 122 1x 121x2 12x 322
and in completely factored form h1x2 1x 122 1x 12 1x 421x 82 . The
ocean floor has height zero at distances of
450
1, 4, 8, and 12 mi from shore.
The graph of h(x) is shown in the figure. The
graph shows a great deal of variation in the
ocean floor, but the zeroes occurring at 1, 4, 8,
and 12 mi out are clearly evident.
E. You’ve just seen how
we can solve an application
of polynomial functions
0
13
450
Now try Exercises 99 through 110
䊳
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4.2 EXERCISES
䊳
CONCEPTS AND VOCABULAR Y
Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.
1. A complex polynomial is one where one or more
are complex numbers.
2. A polynomial function of degree n will have
exactly
zeroes, real or
, where
zeroes of multiplicity m are counted m times.
3. If a bi is a complex zero of polynomial P with
real coefficients, then
is also a zero.
4. According to Descartes’ rule of signs, there are as
many
real roots as changes in sign from
term to term, or an
number less.
5. Which of the following values is not a possible root
of f 1x2 6x3 2x2 5x 12:
a. x 43
b. x 34
c. x 12
6. Discuss/Explain each of the following:
(a) irreducible quadratic factors, (b) factors that are
complex conjugates, (c) zeroes of multiplicity m,
and (d) upper bounds on the zeroes of a
polynomial.
Discuss/Explain why.
䊳
DEVELOPING YOUR SKILLS
Rewrite each polynomial as a product of linear factors,
and find the zeroes of the polynomial.
7. P1x2 x4 5x2 36
8. Q1x2 x4 21x2 100
9. Q1x2 x4 16
19. degree 3, x 3, x 2i
20. degree 3, x 5, x 3i
10. P1x2 x4 81
21. degree 4, x 1, x 2, x i
11. P1x2 x3 x2 x 1
12. Q1x2 x3 3x2 9x 27
13. Q1x2 x3 5x2 25x 125
14. P1x2 x3 4x2 16x 64
Factor each polynomial completely. Write any repeated
factors in exponential form, then name all zeroes and
their multiplicity.
15. p1x2 1x2 10x 252 1x2 4x 452 1x 92
16. q1x2 1x2 12x 362 1x2 2x 242 1x 42
17. P1x2 1x 5x 142 1x 492 1x 22
2
Find a polynomial P(x) having real coefficients,
with the degree and zeroes indicated. All real
zeroes are given. Assume the lead coefficient is 1.
Recall 1a ⴙ bi21a ⴚ bi2 ⴝ a2 ⴙ b2.
2
18. Q1x2 1x2 9x 182 1x2 362 1x 32
22. degree 4, x 1, x 3, x 2i
23. degree 4, x 3, x 2i
24. degree 4, x 2, x 3i
25. degree 4, x 1, x 1 2i
26. degree 4, x 1, x 1 3i
27. degree 4, x 3, x 1 i12
28. degree 4, x 2, x 1 i 13
Use the intermediate value theorem to verify the given
polynomial has at least one zero “ci” in the intervals
specified. Do not find the zeroes.
29. f 1x2 x3 2x2 8x 5
a. 3 4, 34
b. [2, 3]
30. g1x2 x4 2x2 6x 3
a. 33, 24
b. [0, 1]
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Section 4.2 The Zeroes of Polynomial Functions
For Exercises 31 and 32, enter each function on the Y=
screen. Then place your graphing calculator in G-T MODE
and set up a TABLE using TblStart ⴝ ⴚ5 and
¢TBL 0.1. Use the intermediate value theorem and the
resulting GRAPH and TABLE to locate intervals [x1, x2],
where x2 x1 0.1, that contain zeroes of the function.
Assume all real zeroes are between 5 and 5.
31. f 1x2 x3 6x2 4x 10
32. g1x2 2x4 3x3 7x2 5x 4
List all possible rational zeroes for the polynomials
given, but do not solve.
33. f 1x2 4x3 19x 15
34. g1x2 3x3 2x 20
35. h1x2 2x3 5x2 28x 15
36. H1x2 2x3 19x2 37x 14
37. p1x2 6x4 2x3 5x2 28
38. q1x2 7x4 6x3 49x2 36
39. Y1 32t3 52t2 17t 3
40. Y2 24t3 17t2 13t 6
Use the rational zeroes theorem to write each function
in factored form and find all zeroes. Note a ⴝ 1.
56. H1x2 9x3 3x2 8x 4
57. Y1 2x3 3x2 9x 10
58. Y2 3x3 14x2 17x 6
59. p1x2 2x4 3x3 9x2 15x 5
60. q1x2 3x4 x3 11x2 3x 6
61. r1x2 3x4 4x3 8x2 16x 16
62. s1x2 2x4 7x3 14x2 63x 36
Find the zeroes of the polynomials given using any
combination of the rational zeroes theorem, testing for
1 and ⴚ1, and/or the remainder and factor theorems.
63. f 1x2 2x4 9x3 4x2 21x 18
64. g1x2 3x4 4x3 21x2 10x 24
65. h1x2 3x4 2x3 9x2 4
66. H1x2 7x4 6x3 49x2 36
67. p1x2 2x4 3x3 24x2 68x 48
68. q1x2 3x4 19x3 6x2 96x 32
69. r1x2 3x4 20x3 34x2 12x 45
70. s1x2 4x4 15x3 9x2 16x 12
71. Y1 x5 6x2 49x 42
41. f 1x2 x3 13x 12
72. Y2 x5 2x2 9x 6
43. h1x2 x3 19x 30
74. P1x2 2x5 x4 3x3 4x2 14x 12
42. g1x2 x3 21x 20
44. H1x2 x3 28x 48
45. p1x2 x3 2x2 11x 12
46. q1x2 x3 4x2 7x 10
47. Y1 x3 6x2 x 30
48. Y2 x3 4x2 20x 48
49. Y3 x4 15x2 10x 24
50. Y4 x4 23x2 18x 40
51. f 1x2 x4 7x3 7x2 55x 42
52. g1x2 x4 4x3 17x2 24x 36
Find all rational zeroes of the functions given and use
them to write the function in factored form. Use the
factored form to state all zeroes of f. Begin by applying
the tests for 1 and ⴚ1.
53. f 1x2 4x 7x 3
3
54. g1x2 9x 7x 2
3
55. h1x2 4x 8x 3x 9
3
2
333
73. P1x2 3x5 x4 x3 7x2 24x 12
75. Y1 x4 5x3 20x 16
76. Y2 x4 10x3 90x 81
77. r1x2 x4 x3 14x2 2x 24
78. s1x2 x4 3x3 13x2 9x 30
79. p1x2 2x4 x3 3x2 3x 9
80. q1x2 3x4 x3 13x2 5x 10
81. f 1x2 2x5 7x4 13x3 23x2 21x 6
82. g1x2 4x5 3x4 3x3 11x2 27x 6
Gather information on each polynomial using (a) the
rational zeroes theorem, (b) testing for 1 and ⴚ1,
(c) applying Descartes’ rule of signs, and (d) using the
upper and lower bounds property. Respond explicitly
to each.
83. f 1x2 x4 2x3 4x 8
84. g1x2 x4 3x3 7x 6
85. h1x2 x5 x4 3x3 5x 2
86. H1x2 x5 x4 2x3 4x 4
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87. p1x2 x5 3x4 3x3 9x2 4x 12
88. q1x2 x5 2x4 8x3 16x2 7x 14
89. r1x2 2x4 7x2 11x 20
90. s1x2 3x4 8x3 13x 24
Use Descartes’ rule of signs to determine the possible
combinations of real and complex zeroes for each
polynomial. Then graph the function on the standard
window of a graphing calculator and adjust it as needed
until you’re certain all real zeroes are in clear view. Use
䊳
this screen and a list of the possible rational zeroes to
factor the polynomial and find all zeroes (real and
complex).
91. f 1x2 4x3 16x2 9x 36
92. g1x2 6x3 41x2 26x 24
93. h1x2 6x3 73x2 10x 24
94. H1x2 4x3 60x2 53x 42
95. p1x2 4x4 40x3 93x2 30x 72
96. q1x2 4x4 42x3 70x2 21x 36
WORKING WITH FORMULAS
97. The absolute value of a complex number
z ⴝ a ⴙ bi: 円z円 ⴝ 2a2 ⴙ b2
The absolute value of a complex number z, denoted
冟z冟, represents the distance between the origin and
the point (a, b) in the complex plane. Use the
formula to find 冟z冟 for the complex numbers given
(also see Section 3.1, Exercise 69): (a) 3 4i,
(b) 5 12i, and (c) 1 i13.
䊳
4–28
CHAPTER 4 Polynomial and Rational Functions
98. The square root of z ⴝ a ⴙ bi:
1z ⴝ
12
2
1 1 円 z 円 ⴙ a ⴞ i 1 円 z円 ⴚ a2
The principal square root of a complex number is
given by the relation shown, where 冟 z冟 represents
the absolute value of z and the sign for the “ ” is
chosen to match the sign of b. Use the formula to
find the square root of each complex number from
Exercise 97, then check your answer by squaring
the result (also see Section 3.1, Exercise 82).
APPLICATIONS
99. Maximum and minimum values: To locate the
maximum and minimum values of
F 1x2 x4 4x3 12x2 32x 15 requires
finding the zeroes of f 1x2 4x3 12x2 24x 32. Use the rational zeroes theorem and
synthetic division to find the zeroes of f, then graph
F(x) on a calculator and see if the graph tends to
support your calculations — do the maximum and
minimum values occur at the zeroes of f ?
100. Graphical analysis: Use the rational zeroes
theorem and synthetic division to find the zeroes of
F1x2 x4 4x3 12x2 32x 15 (see
Exercise 99 to verify graphically).
101. Maximum and minimum values: To locate the
maximum and minimum values of
G1x2 x4 6x3 x2 24x 20 requires
finding the zeroes of g1x2 4x3 18x2 2x 24.
Use the rational zeroes theorem and synthetic
division to find the zeroes of g, then graph G(x) on
a calculator and see if the graph tends to support
your calculations — do the maximum and
minimum values occur at the zeroes of g?
102. Graphical analysis: Use the rational zeroes
theorem and synthetic division to find the zeroes
of G1x2 x4 6x3 x2 24x 20 (see
Exercise 101 to verify graphically).
Geometry: The volume of a cube is V x # x # x x3,
where x represents the length of the edges. If a slice 1 unit
thick is removed from the cube, the remaining volume is
v x # x # 1x 12 x3 x2. Use this information for
Exercises 103 and 104.
103. A slice 1 unit in thickness is removed from one side
of a cube. Use the rational zeroes theorem and
synthetic division to find the original dimensions of
the cube, if the remaining volume is (a) 48 cm3 and
(b) 100 cm3.
104. A slice 1 unit in thickness is removed from one
side of a cube, then a second slice of the same
thickness is removed from a different side (not the
opposite side). Use the rational zeroes theorem and
synthetic division to find the original dimensions of
the cube, if the remaining volume is (a) 36 cm3 and
(b) 80 cm3.
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Geometry: The volume of a rectangular box is
V LWH. For the box to satisfy certain requirements, its
length must be twice the width, and its height must be two
inches less than the width. Use this information for
Exercises 105 and 106.
105. Use the rational zeroes theorem and synthetic
division to find the dimensions of the box if it must
have a volume of 150 in3.
106. Suppose the box must have a volume of 64 in3. Use
the rational zeroes theorem and synthetic division
to find the dimensions required.
Government deficits: Over a 14-yr period, the balance
of payments (deficit versus surplus) for a certain county
government was modeled by the function
f 1x2 14x4 6x3 42x2 72x 64, where x 0
corresponds to 1990 and f(x) is the deficit or surplus in
tens of thousands of dollars. Use this information for
Exercises 107 and 108.
107. Use the rational zeroes theorem and synthetic
division to find the years when the county “broke
even” (debt surplus 0) from 1990 to 2004.
How many years did the county run a surplus
during this period?
108. The deficit was at the $84,000 level 3 f 1x2 844 ,
four times from 1990 to 2004. Given this occurred
in 1992 and 2000 (x 2 and x 10), use the
rational zeroes theorem, synthetic division, and the
remainder theorem to find the other two years the
deficit was at $84,000.
109. Drag resistance
on a boat: In a
scientific study on
the effects of drag
against the hull of
a sculling boat,
some of the
factors to consider
are displacement, draft, speed, hull shape, and
length, among others. If the first four are held
䊳
Section 4.2 The Zeroes of Polynomial Functions
335
constant and we assume a flat, calm water surface,
length becomes the sole variable (as length
changes, we adjust the beam by a uniform scaling
to keep a constant displacement). For a fixed
sculling speed of 5.5 knots, the relationship
between drag and length can be modeled by
f 1x2 0.4192x4 18.9663x3 319.9714x2 2384.2x 6615.8, where f (x) is the efficiency
rating of a boat with length x (8.7 6 x 6 13.6).
Here, f 1x2 0 represents an average efficiency
rating. (a) Under these conditions, what lengths (to
the nearest hundredth) will give the boat an average
rating? (b) What length will maximize the
efficiency of the boat? What is this rating?
110. Comparing densities:
Why is it that when
you throw a rock into
a lake, it sinks, while a
wooden ball will float
half submerged, but
the bobber on your
fishing line floats on
the surface? It all
depends on the density
of the object compared to the density of water
(d 1). For uniformity, we’ll consider spherical
objects of various densities, each with a radius
of 5 cm. When placed into water, the depth that
the sphere will sink beneath the surface (while
still floating) is modeled by the polynomial
p1x2 ␲3 x3 5␲x2 500␲
3 d, where d is the
density of the object and the smallest positive
zero of p is the depth of the sphere below the
surface (in centimeters). How far submerged is
the sphere if it’s made of (a) balsa wood, d 0.17;
(b) pine wood, d 0.55; (c) ebony wood,
d 1.12; (d) a large bobber made of lightweight
plastic, d 0.05 (see figure)?
EXTENDING THE CONCEPT
111. In the figure, P1x2 0.02x3 0.24x2 1.04x 2.68 is graphed on the standard
screen (10 x 10), which shows two real zeroes. Since P has degree 3,
there must be one more real zero but is it negative or positive? Use the
upper/lower bounds property (a) to see if 10 is a lower bound and (b) to see
if 10 is an upper bound. (c) Then use your calculator to find the remaining zero.
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112. From Example 11, (a) what is the significance of
the y-intercept? (b) If the domain were extended to
include 0 6 x 13, what happens when x is
approximately 12.8?
113A. It is often said that while the difference of two
squares is factorable, a2 b2 1a b2 1a b2,
the sum of two squares is prime. To be 100%
correct, we should say the sum of two squares
cannot be factored using real numbers. If
complex numbers are used,
1a2 b2 2 1a bi2 1a bi2 . Use this idea to
factor the following binomials.
a. p1x2 x2 25
b. q1x2 x2 9
c. r1x2 x2 7
113B. It is often said that while x2 16 is factorable as
a difference of squares,
a2 b2 1a b2 1a b2, x2 17 is not. To be
100% correct, we should say that x2 17 is not
factorable using integers. Since 1 1172 2 17, it
can actually be factored in the same way:
x2 17 1x 1172 1x 1172 . Use this idea
to solve the following equations.
a. x2 7 0
b. x2 12 0
c. x2 18 0
114. Every general cubic equation aw3 bw2 cw d 0 can be written in the form x3 px q 0 (where the squared term has been
“depressed”), using the transformation
b
w x . Use this transformation to solve the
3a
following equations.
a. w3 3w2 6w 4 0
b. w3 6w2 21w 26 0
䊳
4–30
CHAPTER 4 Polynomial and Rational Functions
Note: It is actually very rare that the transformation
produces a value of q 0 for the “depressed”
cubic x3 px q 0, and general solutions must
be found using what has become known as
Cardano’s formula. For a complete treatment of
cubic equations and their solutions, visit our
website at www.mhhe.com/coburn.
115. For each of the following complex polynomials,
one of its zeroes is given. Use this zero to help
write the polynomial in completely factored form.
(Hint: Synthetic division and the quadratic formula
can be applied to all polynomials, even those with
complex coefficients.)
a. C1z2 z3 11 4i2z2 16 4i2z 24i;
z 4i
b. C1z2 z3 15 9i2z2 14 45i2z 36i;
z 9i
c. C1z2 z3 12 3i2z2 15 6i2z 15i;
z 3i
d. C1z2 z3 14 i2z2 129 4i2z 29i;
zi
e. C1z2 z3 12 6i2z2 14 12i2z 24i;
z 6i
f. C1z2 z3 16 4i2z2 111 24i2 z 44i;
z 4i
g. C1z2 z3 12 i2z2 15 4i2z 16 3i2;
z2i
h. C1z2 z3 2z2 119 6i2z 120 30i2;
z 2 3i
MAINTAINING YOUR SKILLS
116. (2.5) Graph the piecewise-defined function and
find the values of f 132, f (2), and f (5).
2
f 1x2 • 冟 x 1冟
4
x 1
1 6 x 6 5
x5
117. (3.4) For a county fair, officials need to fence off a
large rectangular area, then subdivide it into three
equal (rectangular) areas. If the county provides
1200 ft of fencing, (a) what dimensions will
maximize the area of the larger (outer) rectangle?
(b) What is the area of each smaller rectangle?
118. (2.1) Use the graph given
to (a) state intervals where
f 1x2 0, (b) locate local
maximum and minimum
values, and (c) state
intervals where f 1x2c
and f 1x2T.
y
5
f (x)
5
5 x
5
y
119. (2.2) Write the equation of the
function shown.
5
5
r(x)
5 x
5
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Graphing Polynomial Functions
LEARNING OBJECTIVES
In Section 4.3 you will see
how we can:
A. Identify the graph of a
B.
C.
D.
E.
polynomial function and
determine its degree
Describe the endbehavior of a polynomial
graph
Discuss the attributes of
a polynomial graph with
zeroes of multiplicity
Graph polynomial
functions in standard
form
Solve applications of
polynomials and
polynomial modeling
As with linear and quadratic functions, understanding graphs of polynomial functions
will help us apply them more effectively as mathematical models. Since all real polynomials can be written in terms of their linear and quadratic factors (Section 4.2), these
functions provide the basis for our continuing study.
A. Identifying the Graph of a Polynomial Function
Figure 4.9
Consider the graphs of f 1x2 x 2 and
y
P(x)
g1x2 1x 12 2, which we know are smooth, continuous
5
curves. The graph of f is a straight line with positive
4
3
slope, that crosses the x-axis at 2. The graph of g is a
2
parabola, opening upward, shifted 1 unit to the right,
1
and touching the x-axis at x 1. When f and g are “com2
bined” into the single function P1x2 1x 221x 12 , 5 4 3 2 11 1 2 3 4 5 x
the behavior of the graph at these zeroes is still evident.
2
3
In Figure 4.9, the graph of P crosses the x-axis at
4
x 2, “bounces” off the x-axis at x 1, and is still
5
a smooth, continuous curve. This observation could
be extended to include additional linear or quadratic
factors, and helps affirm that the graph of a polynomial function is a smooth, continuous curve.
Further, after the graph of P crosses the axis at x 2, it must “turn around” at some
point to reach the zero at x 1, then turn again as it touches the x-axis without crossing.
By combining this observation with our work in Section 4.2, we can state the following:
Polynomial Graphs and Turning Points
1. If P(x) is a polynomial function of degree n,
then the graph of P has at most n 1 turning points.
2. If the graph of a function P has n 1 turning points,
then the degree of P(x) is at least n.
While defined more precisely in a future course, we will take “smooth” to mean
the graph has no sharp turns or jagged edges, and “continuous” to mean the entire
graph can be drawn without lifting your pencil (Figure 4.10). In other words, a polynomial graph has none of the attributes shown in Figure 4.11.
Figure 4.10
Figure 4.11
cusp
gap
hole
sharp turn
polynomial
4–31
nonpolynomial
337
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College Algebra
Graphs & Models—
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4–32
CHAPTER 4 Polynomial and Rational Functions
EXAMPLE 1
䊳
Identifying Polynomial Graphs
Determine whether each graph could be the graph of a polynomial. If not, discuss
why. If so, use the number of turning points and zeroes to identify the least
possible degree of the function.
a.
b.
y
y
5
5
4
4
3
3
2
2
1
1
5 4 3 2 1
1
A. You’ve just seen how
we can identify the graph of a
polynomial function and
determine its degree
3
4
5
5 4 3 2 1
1
x
3
3
4
4
5
5
d.
y
5
4
4
3
3
2
2
1
1
1
2
3
4
5
x
1
2
3
4
5
x
2
3
4
5
x
y
5
5 4 3 2 1
1
䊳
2
2
c.
Solution
1
2
5 4 3 2 1
1
2
2
3
3
4
4
5
5
1
a. This is not a polynomial graph, as it has a cusp at (1, 3).
A polynomial graph is always smooth.
b. This graph is smooth and continuous, and could be that of a polynomial.
With two turning points and three zeroes, the function is at least degree 3.
c. This graph is smooth and continuous, and could be that of a polynomial.
With three turning points and two zeroes, the function is at least degree 4.
d. This is not a polynomial graph, as it has a gap (discontinuity) at x 1.
A polynomial graph is always continuous.
Now try Exercises 7 through 12
䊳
B. The End-Behavior of a Polynomial Graph
Once the graph of a function has “made its last turn” and crossed or touched its last real
zero, it will continue to increase or decrease without bound as 冟x冟 becomes large. As
before, we refer to this as the end-behavior of the graph. In previous sections, we noted
that quadratic functions (degree 2) with a positive leading coefficient 1a 7 02, had the
end-behavior “up on the left” and “up on the right (up/up).” If the leading coefficient
was negative 1a 6 02, end-behavior was “down on the left” and “down on the right
(down/down).” These descriptions were also applied to the graph of a linear function
y mx b (degree 1). A positive leading coefficient 1m 7 02 indicates the graph will
be down on the left, up on the right (down/up), and so on. All polynomial graphs exhibit some form of end-behavior, which can be likewise described.
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Section 4.3 Graphing Polynomial Functions
EXAMPLE 2
䊳
Identifying the End-Behavior of a Graph
State the end-behavior of each graph shown:
a. f 1x2 x3 4x 1
b. g1x2 2x5 7x3 4x
y
䊳
y
y
5
5
5
4
4
4
3
3
3
2
2
2
1
1
1
5 4 3 2 1
1
Solution
c. h1x2 2x4 5x2 x 1
1
2
3
4
5
x
5 4 3 2 1
1
1
2
3
4
5
x
5 4 3 2 1
1
2
2
2
3
3
3
4
4
4
5
5
5
a. down on the left, up on the right
c. down on the left, down on the right
1
2
3
4
5
b. up on the left, down on the right
Now try Exercises 13 through 16
WORTHY OF NOTE
As a visual aid to end-behavior, it
might help to picture a signalman
using semaphore code as
illustrated here. As you view the
end-behavior of a polynomial
graph, there is a striking
resemblance.
Figure 4.12
down/down
up/down
down/up
䊳
The leading term ax n of a polynomial function is said to be the dominant term,
because for large values of 冟x冟, the value of ax n is much larger than all other terms combined. Figure 4.12 shows a table of values for Y1 0.4x5 3x4 9x3 15x2 30,
where the leading coefficient is positive and very small, with all other coefficients
negative and much larger. Initially, all outputs are negative as these terms overpower
the leading term. But eventually (in this case for any integer greater than 10), the leading term will dominate all others since it becomes much more “powerful” for larger
values. See Figure 4.13.
Y1 0.4x5 3x4 9x3 15x2 30
up/up
x
Figure 4.13
This means that like linear and quadratic graphs, polynomial end-behavior can be
predicted in advance by analyzing this term alone.
1. For axn when n is even, any nonzero number raised to an even power is positive,
so the ends of the graph must point in the same direction. If a 7 0, both point upward. If a 6 0, both point downward.
2. For ax n when n is odd, any number raised to an odd power has the same sign as the
input value, so the ends of the graph must point in opposite directions. If a 7 0,
end-behavior is down on the left, up on the right. If a 6 0, end-behavior is up on
the left, down on the right.
From this we find that end-behavior depends on two things: the degree of the function
(even or odd) and the sign of the leading coefficient (positive or negative). In more formal
terms, this is described in terms of how the graph “behaves” for large values of x. For endbehavior that is “up on the right,” we mean that as x becomes a large positive number, y
becomes a large positive number. This is indicated using the notation: as x S q, y S q.
Similar notation is used for the other possibilities. These facts are summarized in
Table 4.1. The interior portion of each graph is dashed since the actual number of turning
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CHAPTER 4 Polynomial and Rational Functions
points may vary, although a polynomial of odd degree will have an even number of turning points, and a polynomial of even degree will have an odd number of turning points.
Table 4.1
Polynomial End-Behavior
Even Degree:
a⬎0
as x → q, y → q
a⬍0
y
as x → q, y → q
y
x
x
as x → q, y → q
up on the left, up on the right
(similar to y x2)
Odd Degree:
down on the left, down on the right
(similar to y x2)
a⬎0
y
as x → q, y → q
a⬍0
as x → q, y → q
as x → q, y → q
y
x
x
as x → q, y → q
as x → q, y → q
down on the left, up on the right
(similar to y x)
up on the left, down on the right
(similar to y x)
Note the end-behavior of y mx can be used as a representative of all odd degree functions, and the end-behavior of y ax2 as a representative of all even degree functions.
The End-Behavior of a Polynomial Graph
Given a polynomial P(x) with leading term axn and n 1.
If n is even, ends will point in the same direction,
1. for a 7 0: up on the left, up on the right (as with y x2);
as x S q, y S q;
as x S q, y S q
2. for a 6 0: down on the left, down on the right (as with y x2);
as x S q, y S q;
as x S q, y S q
If n is odd, the ends will point in opposite directions,
1. for a 7 0: down on the left, up on the right (as with y x);
as x S q, y S q;
as x S q, y S q
2. for a 6 0: up on the left, down on the right (as with y x);
as x S q, y S q;
as x S q, y S q
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Section 4.3 Graphing Polynomial Functions
EXAMPLE 3
Solution
䊳
Identifying the End-Behavior of a Function
State the end-behavior of each function.
a. f 1x2 0.8x4 3x3 0.5x2 4x 1
䊳
b. g1x2 2x5 6x3 1
a. The function has degree 4 (even), so the ends will point in the same direction.
The leading coefficient is positive, so end-behavior is up/up. See Figure 4.14.
b. The function has degree 5 (odd), so the ends will point in opposite directions.
The leading coefficient is negative, so the end-behavior is up/down. See
Figure 4.15.
Figure 4.14
Figure 4.15
5
10
3
5
4
4
5
B. You’ve just seen how
we can describe the endbehavior of a polynomial graph
10
Now try Exercises 17 through 22
䊳
C. Attributes of Polynomial Graphs with Zeroes of Multiplicity
Another important aspect of polynomial functions is the behavior of a graph near its
zeroes. In the simplest case, consider the functions f 1x2 x and g1x2 x3 in Figure 4.16. Both have odd degree, like end-behavior (down/up), and a zero at x 0.
But the zero of f has multiplicity 1, while the zero from g has multiplicity 3. Notice
the graph of g is vertically compressed near x 0 and flattens out on its approach
and departure from this zero.
Figure 4.16
5
y g(x) x3
2
1
5
5
x
2
1
1
2
1
f(x) x
2
5
This behavior can be explained by noting that for x 1 and 1, f 1x2 g1x2. But for
冟x冟 6 1, the graph of g will be closer to the x-axis (g decreases faster than f ) since the
cube of a fractional number is smaller than the fraction itself. We further note that for
冟x冟 7 1, g increases much faster than f, and 0 g1x2 0 7 0 f 1x2 0 . Similar observations can
be made regarding f 1x2 x2 and g1x2 x4 in Figure 4.17. Both functions have even
degree, a zero at x 0, and f 1x2 g1x2 for x 1 and 1. But for 冟x冟 6 1, the function with higher degree is once again closer to the x-axis.
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CHAPTER 4 Polynomial and Rational Functions
Figure 4.17
5
y g(x) x4
2
f(x) x2
1
5
5
2
x
1
1
2
1
2
5
These observations can be generalized and applied to all real zeroes of a function.
Polynomial Graphs and Zeroes of Multiplicity
Given P(x) is a polynomial with factors of the form 1x c2 m, with c a real number,
• If m is odd, the graph will cross through the x-axis.
• If m is even, the graph will “bounce” off the x-axis (touching at just one point).
In each case, the graph will be more compressed (flatter) near c for larger values of m.
To illustrate, compare the graph of P1x2 1x 22 1x 12 2 (Figure 4.18), with the
graph of p1x2 1x 22 3 1x 12 4 shown in Figure 4.19, noting the increased multiplicity of each zero.
Figure 4.18
5
Figure 4.19
y P(x)
5
4
4
3
3
2
2
1
1
5 4 3 2 1
1
1
2
3
4
5
x
5 4 3 2 1
1
2
2
3
3
4
4
5
5
y p(x)
1
2
3
4
5
x
Both graphs show the expected zeroes at x 2 and x 1, but the graph of p(x)
is flatter near x 2 and x 1, due to the increased multiplicity of each zero. We
also lose sight of the graph of p(x) between x 2 and x 0, since the increased
multiplicities produce larger values than the original grid could display.
EXAMPLE 4
䊳
Naming Attributes of a Function from Its Graph
The graph of a polynomial f 1x2 is shown.
a. State whether the degree of f is even or odd.
b. Use the graph to name the zeroes of f, then
state whether their multiplicity is even or odd.
c. State the minimum possible degree of f.
d. State the domain and range of f.
y
12
f(x)
10
8
6
4
2
5 4 3 2 1
2
4
6
8
1
2
3
4
5
x
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Section 4.3 Graphing Polynomial Functions
Solution
䊳
a. Since the ends of the graph point in opposite directions, the degree of the
function must be odd.
b. The graph crosses the x-axis at x 3 and is compressed near 3, meaning it
must have odd multiplicity with m 7 1. The graph “bounces” off the x-axis at
x 2 and 2 must be a zero of even multiplicity.
c. The minimum possible degree of f is 5, as in f 1x2 a1x 22 2 1x 32 3.
d. x 僆 ⺢, y 僆 ⺢.
Now try Exercises 23 through 28
䊳
To find the degree of a polynomial from its factored form, add the exponents on all linear factors, then add 2 for each irreducible quadratic factor (the degree of any quadratic
factor is 2). The sum gives the degree of the polynomial, from which end-behavior can
be determined. To find the y-intercept, substitute 0 for x as before, noting this is equivalent to applying the exponent to the constant from each factor.
EXAMPLE 5
Solution
䊳
Naming Attributes of a Function from Its F actored Form
State the degree of each function, then describe the end-behavior and name the
y-intercept of each graph.
a. f 1x2 1x 22 3 1x 32
b. g1x2 1x 22 2 1x2 521x 52
䊳
a. The degree of f is 3 1 4. With even degree and positive leading
coefficient, end-behavior is up/up. For f 102 122 3 132 24, the y-intercept
is 10, 242. See Figure 4.20.
b. The degree of g is 2 2 1 5. With odd degree and negative leading
coefficient, end-behavior is up/down. For g102 1122 2 152152 100, the
y-intercept is (0, 100). See Figure 4.21.
Figure 4.20
Figure 4.21
100
5
1000
5
100
5
7
200
Now try Exercises 29 through 36
EXAMPLE 6
䊳
䊳
Matching Graphs to Functions Using Zeroes of Multiplicity
The following functions all have zeroes at x 2, 1, and 1. Match each function
to the corresponding graph using its degree and the multiplicity of each zero.
a. y 1x 22 1x 12 2 1x 12 3
b. y 1x 22 1x 121x 12 3
c. y 1x 22 2 1x 12 2 1x 12 3
d. y 1x 22 2 1x 121x 12 3
Solution
䊳
The functions in Figures 4.22 and 4.24 must have even degree due to end-behavior,
so each corresponds to (a) or (d). At x 1 the graph in Figure 4.22 “crosses,”
while the graph in Figure 4.24 “bounces.” This indicates Figure 4.22 matches
equation (d), while Figure 4.24 matches equation (a).
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The graphs in Figures 4.23 and 4.25 must have odd degree due to end-behavior, so
each corresponds to (b) or (c). Here, one graph “bounces” at x 2, while the other
“crosses.” The graph in Figure 4.23 matches equation (c), the graph in Figure 4.25
matches equation (b).
Figure 4.22
Figure 4.23
y
2
1
y
5
5
4
4
3
3
2
2
1
1
1
1
2
2
x
1
2
2
3
3
4
4
5
5
Figure 4.24
1
x
2
x
y
5
5
4
4
3
3
2
2
1
1
1
2
Figure 4.25
y
2
1
1
1
2
2
x
1
1
1
2
2
3
3
4
4
5
5
Now try Exercises 37 through 42
䊳
Using the ideas from Examples 5 and 6, we’re able to draw a fairly accurate graph
given the factored form of a polynomial. Convenient values between two zeroes, called
midinterval points, should be used to help complete the graph.
EXAMPLE 7
䊳
Graphing a Function Given the F actored Form
Solution
䊳
Adding the exponents of each factor, we find
that f is a function of degree 6 with a positive
lead coefficient, so end-behavior will be up/up.
Since f 102 2, the y-intercept is 10, 22. The
graph will bounce off the x-axis at x 1 (even
multiplicity), and cross the axis at x 1 and 2
(odd multiplicities). The graph will “flatten out”
near x 1 because of its higher multiplicity.
To help “round-out” the graph we evaluate f at
x 1.5, giving 10.52 10.52 2 12.52 3 ⬇ 1.95
(note scaling of the x- and y-axes).
C. You’ve just seen how
we can discuss the attributes
of a polynomial graph with
zeroes of multiplicity
Sketch the graph of f 1x2 1x 22 1x 12 2 1x 12 3 using end-behavior; the
x- and y-intercepts, and zeroes of multiplicity.
y
5
(1, 0)
(1, 0)
3
2
1
1
(2, 0)
2
3
x
(0, 2)
(1.5, 1.95)
5
Now try Exercises 43 through 56
䊳
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Section 4.3 Graphing Polynomial Functions
D. The Graph of a Polynomial Function
Using the cumulative observations from this and previous sections, a general strategy
emerges for the graphing of polynomial functions.
Guidelines for Graphing Polynomial Functions
WORTHY OF NOTE
Although of somewhat limited value,
symmetry (item f in the guidelines)
can sometimes aid in the graphing
of polynomial functions. If all terms
of the function have even degree,
the graph will be symmetric to the
y-axis (even). If all terms have odd
degree, the graph will be symmetric
to the origin. Recall that a constant
term has degree zero, an even
number.
EXAMPLE 8
䊳
1. Determine the end-behavior of the graph.
2. Find the y-intercept 10, a0 2
3. Find the zeroes using any combination of the rational zeroes theorem, the
factor and remainder theorems, tests for 1 and 1, factoring, and
the quadratic formula.
4. Use the y-intercept, end-behavior, the multiplicity of each zero, and
midinterval points as needed to sketch a smooth, continuous curve.
Additional tools include (a) polynomial zeroes theorem, (b) complex conjugates
theorem, (c) number of turning points, (d) Descartes’ rule of signs, (e) upper and
lower bounds, and (f) symmetry.
Graphing a Polynomial Function
Sketch the graph of g1x2 x4 9x2 4x 12.
Solution
䊳
1. End-behavior: The function has degree 4 (even) with a negative leading
coefficient, so end-behavior is down on the left, down on the right.
2. Since g102 12, the y-intercept is 10, 122.
3. Zeroes: Using the test for x 1 gives 1 9 4 12 8, showing
x 1 is not a zero but 11, 82 is a point on the graph. Using the test for
x 1 gives 1 9 4 12 0, so 1 is a zero and 1x 12 is a factor.
Using x 1 with the factor theorem yields
1|
1
1
0
1
1
9
1
8
4
8
12
12
12
|0
The quotient polynomial is not easily factorable so we continue with synthetic
division. Using the rational zeroes theorem, the possible rational zeroes are
51, 12, 2, 6, 3, 46, so we try x 2.
use 2 as a “divisor” on
the quotient polynomial
2|
1
1
1
2
1
This shows x 2 is a zero, x 2 is a factor,
and the function can now be written as
g1x2 1x 12 1x 22 1x2 x 62.
Factoring 1 from the trinomial gives
g1x2 11x 12 1x 22 1x2 x 62
11x 12 1x 22 1x 32 1x 22
11x 12 1x 22 2 1x 32
The zeroes of g are x 1 and 3, both with
multiplicity 1, and x 2 with multiplicity 2.
12
12
|0
8
2
6
y
g(x)
20
(2, 16)
(2, 0)
(3, 0)
5
5
(1, 0)
(0, 12) (1, 8)
Down on
left
20
Down on
right
x
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CHAPTER 4 Polynomial and Rational Functions
4. To help “round-out” the graph we evaluate the midinterval point x 2 using
the remainder theorem or the factored form of g(x), which shows that 12, 162
is also a point on the graph.
use 2 as a “divisor”
2|
1
0
2
2
1
9
4
5
4
10
14
12
28
| 16
The final result is the graph shown.
Now try Exercises 57 through 72
CAUTION
EXAMPLE 9
䊳
䊳
䊳
Sometimes using a midinterval point to help draw a graph will give the illusion that a
maximum or minimum value has been located. This is rarely the case, as demonstrated
in the figure in Example 8, where the maximum value in Quadrant II is actually closer to
12.22, 16.952.
Using the Guidelines to Sk etch a Polynomial Graph
Sketch the graph of h1x2 x7 4x6 7x5 12x4 12x3.
Solution
䊳
1. End-behavior: The function has degree 7 (odd) so the ends will point in
opposite directions. The leading coefficient is positive and the end-behavior
will be down on the left and up on the right.
2. y-intercept: Since h102 0, the y-intercept is (0, 0).
3. Zeroes: Testing 1 and 1 shows neither are zeroes but (1, 4) and 11, 362
are points on the graph. Factoring out x3 produces h1x2 x3 1x4 4x3 7x2 12x 122, and we see that x 0 is a zero of multiplicity 3. We next use
synthetic division with x 2 on the fourth-degree polynomial:
use 2 as a “divisor”
2|
1
1
4
2
2
This shows x 2 is a zero and x 2 is a
factor. At this stage, it appears the quotient
can be factored by grouping. From
h1x2 x3 1x 22 1x3 2x2 3x 62, we
obtain h1x2 x3 1x 22 1x2 32 1x 22
after factoring and
h1x2 x 1x 22 1x 32
3
2
7
4
3
12
6
6
12
12
|0
y
10
(1, 4)
h(x)
Up on
right
(0, 0)
3
(2, 0)
3
x
2
Down on
as the completely factored form. We find
left
10
that x 2 is a zero of multiplicity 2, and
the remaining two zeroes are complex.
4. Using this information produces the graph shown in the figure.
Now try Exercises 73 through 76
D. You’ve just seen how
we can graph polynomial
functions in standard form
䊳
For practice with these ideas using a graphing calculator, see Exercises 77 through 80.
Similar to our work in previous sections, Exercises 81 and 82 ask you to reconstruct
the complete equation of a polynomial from its given graph.
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Section 4.3 Graphing Polynomial Functions
347
E. Applications of Polynomials and Polynomial Modeling
EXAMPLE 10
䊳
Modeling the Value of an Investment
In the year 2000, Marc and his wife Maria decided to invest some money in precious
metals. As expected, the value of the investment fluctuated over the years, sometimes
being worth more than they paid, other times less. Suppose that through 2010, the gain
or loss on the investment was modeled by v1t2 t 4 11t 3 38t 2 40t, where v(t)
represents the gain or loss (in hundreds of dollars) in year t 1t 0 S 20002.
a. Use the rational zeroes theorem to find the years when their gain/loss was zero.
b. Sketch the graph of the function.
c. In what years was the investment worth less than they paid?
d. What was their gain or loss in 2010?
Solution
䊳
WORTHY OF NOTE
Due to the context, the domain of
v(t) in Example 10 actually begins
at t 0, which we could designate
with a point at (0, 0). In addition,
note there are three sign changes in
the terms of v(t), indicating there
will be 3 or 1 positive roots (we
found 3).
a. Writing the function as v1t2 t1t3 11t2 38t 402, we note t 0 shows
no gain or loss on purchase, and attempt to find the remaining zeroes. Testing
for 1 and 1 shows neither is a zero, but 11, 122 and 11, 902 are points on
the graph of v. Next we try t 2 with the factor theorem and the cubic
polynomial.
1
1
1
10
(3, 6)
x
10|
1
1
(1, 12)
40
40
|0
11
3
8
38
24
14
40
42
2
0
6
|6
The graph is shown in Figure 4.26.
c. The investment was worth less than what they paid (outputs are negative) from
2000 to 2002 and 2004 to 2005.
d. In 2010, they were “sitting pretty,” as their
Figure 4.27
investment had gained $2,400.
Figure 4.26
y
10
38
18
20
We find that 2 is a zero and write v1t2 t1t 22 1t2 9t 202, then factor to
obtain v1t2 t1t 22 1t 42 1t 52. Since v1t2 0 for t 0, 2, 4, and 5, they
“broke even” in years 2000, 2002, 2004, and 2005.
b. With even degree and a positive leading coefficient, the end-behavior is up/up.
All zeroes have multiplicity 1. As an additional midinterval point we find
v132 6:
3|
5
11
2
9
1
2|
11
10
1
38
10
28
40
280
240
0
2400
|2400
These values can also be calculated or
confirmed using a graphing calculator.
See Figure 4.27.
Now try Exercises 85 through 88
䊳
As with linear and quadratic regression models, applications of other polynomial
models begins with a scatterplot and a decision as to which form of regression might
be appropriate. This can depend on a number of factors, such as any end-behavior that is
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CHAPTER 4 Polynomial and Rational Functions
evident, the number of apparent turning points, any anticipated behavior, and so on.
However, due to the end-behavior of polynomial models, great care must be exercised
when these models are used to make projections beyond the limits of the given data.
EXAMPLE 11
Solution
䊳
䊳
The Earth’s atmosphere consists of several layers that
are defined in terms of altitude and the characteristics
of the air in each layer. In order, these are the
Troposphere (0–12 km), Stratosphere (12–50 km),
Mesosphere (50–80 km), and Thermosphere
(80–100 km). Due to their chemical and physical
characteristics, the air temperature within each layer
and from layer to layer varies a great deal. The data
in the table gives the temperature in °C at an altitude
of h kilometers (km). Use the data to:
a. Draw a scatterplot and decide on an appropriate
form of regression, then find the regression
equation.
b. Use the regression equation to find the
temperature at altitudes of 32.6 km and 63.6 km.
c. As the space shuttle rockets into orbit, a
temperature reading of 75°C is taken. What are
the possible altitudes for the shuttle at this point?
Altitude
(km)
Temperature
(ⴗC)
0
20
4
20
8
45
12
55
20
57
30
43
40
16
50
2
60
14
70
54
80
91
90
93
100
45
a. The scatterplot is shown in Figure 4.28. Using the characteristics exhibited
(end-behavior, three turning points), it appears a quartic regression (degree 4)
is appropriate and the equation is shown in Figure 4.29.
Figure 4.28
Figure 4.29
40
10
110
110
b. At altitudes of 32.6 km and 63.6 km, the temperature is very near 30.0°C
(Figure 4.30).
c. Setting Y2 75, we note the line intersects the graph in two places, one
indicating an altitude of about 76.4 km, and the other an altitude of about
96.1 km (Figure 4.31).
Figure 4.31
Figure 4.30
40
10
E. You’ve just seen how
to solve applications of
polynomials and polynomial
modeling
110
110
Now try Exercises 89 through 92
䊳
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Section 4.3 Graphing Polynomial Functions
4.3 EXERCISES
䊳
CONCEPTS AND VOCABULAR Y
Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.
1. For a polynomial with factors of the form 1x c2 m,
c is called a
of multiplicity
.
3. The graphs of Y1 1x 22 2 and Y2 1x 22 4
both
at x 2, but the graph of Y2 is
than the graph of Y1 at this point.
5. In your own words, explain/discuss how to find the
degree and y-intercept of a function that is given in
factored form. Use f 1x2 1x 12 3 1x 22 1x 42 2
to illustrate.
䊳
2. A polynomial function of degree n has
zeroes and at most
“turning points.”
4. Since x4 7 0 for all x, the ends of its graph will
always point in the
direction. Since
x3 7 0 when x 7 0 and x3 6 0 when x 6 0,
the ends of its graph will always point in the
direction.
6. Name all of the “tools” at your disposal that play
a role in the graphing of polynomial functions.
Which tools are indispensable and always used?
Which tools are used only as the situation merits?
DEVELOPING YOUR SKILLS
Determine whether each graph is the graph of a
polynomial function. If yes, state the least possible
degree of the function. If no, state why.
7.
y
f(x)
12
10
8
6
4
2
54321
2
4
6
8
8.
y
1 2 3 4 5 x
13. f (x)
1 2 3 4 5 x
14. g(x)
y
10
8
6
4
2
h(x)
5
4
3
2
1
54321
1
2
3
4
5
State the end-behavior of the functions shown.
54321
2
4
6
8
10
y
g(x)
5
4
3
2
1
54 321
1
10.
30
24
18
12
6
1 2 3 4 5 x
y
q(x)
1 2 3 4 5 x
12.
y
5
4
3
2
1
54321
1
2
3
4
5
f(x)
1 2 3 4 5 x
54321
6
12
18
24
30
1 2 3 4 5 x
16. h(x)
y
p(x)
2
3
4
5
5
4
3
2
1
54321
1
2
3
4
5
5
4
3
2
1
54 321
1
1 2 3 4 5 x
2
3
4
5
11.
y
g(x)
30
24
18
12
6
1 2 3 4 5 x
15. H(x)
9.
y
f(x)
54321
6
12
18
24
30
y
H(x)
1 2 3 4 5 x
h(x)
325
260
195
130
65
108642
65
2 4 6 8 10 x
130
195
260
325
State the end-behavior and y-intercept of the functions
given. Do not graph.
17. f 1x2 x3 6x2 5x 2
18. g1x2 x4 4x3 2x2 16x 12
19. p1x2 2x4 x3 7x2 x 6
20. q1x2 2x3 18x2 7x 3
21. Y1 3x5 x3 7x2 6
22. Y2 x6 4x5 4x3 16x 12
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For each polynomial graph, (a) state whether the degree
of the function is even or odd; (b) use the graph to name
the zeroes of f, then state whether their multiplicity is
even or odd; (c) state the minimum possible degree of f
and write one possible function for f in factored form;
and (d) estimate the domain and range. Assume all
zeroes are real.
23.
54321
2
4
6
8
10
1 2 3 4 5 x
40
32
24
16
8
54321
8
16
24
32
40
c.
e.
y
1 2 3 4 5 x
State the degree of each function, the end-behavior, and
y-intercept of its graph.
29. f 1x2 1x 32 1x 12 3 1x 22 2
30. g1x2 1x 22 1x 42 1x 12
2
31. Y1 1x 12 2 1x 22 12x 32 1x 42
32. Y2 1x 12 1x 22 3 15x 32
33. r1x2 1x2 32 1x 42 3 1x 12
34. s1x2 1x 22 2 1x 12 2 1x2 52
35. h1x2 1x2 22 1x 12 2 11 x2
36. H1x2 1x 22 2 12 x2 1x2 42
Every function in Exercises 37 through 42 has the zeroes
x ⴝ ⴚ1, x ⴝ ⴚ3, and x ⴝ 2. Match each to its
corresponding graph using degree, end-behavior, and
the multiplicity of each zero.
37. f 1x2 1x 12 2 1x 32 1x 22
38. F1x2 1x 12 1x 32 2 1x 22
39. g1x2 1x 12 1x 32 1x 22 3
54321
4
1 2 3 4 5 x
1 2 3 4 5 x
8
12
16
20
y
25
20
15
10
5
54321
5
10
15
20
25
1 2 3 4 5 x
20
16
12
8
4
54321
4
8
12
16
20
y
20
16
12
8
4
1 2 3 4 5 x
y
28.
y
10
8
6
4
2
42. Y2 1x 12 3 1x 32 1x 22 2
y
a.
b.
60
48
36
24
12
54321
12
24
36
48
60
1 2 3 4 5 x
41. Y1 1x 12 2 1x 32 1x 22 2
y
26.
y
40. G1x2 1x 12 3 1x 32 1x 22
10
8
6
4
2
54321
2
4
6
8
10
1 2 3 4 5 x
30
24
18
12
6
54321
6
12
18
24
30
27.
24.
y
10
8
6
4
2
54321
2
4
6
8
10
25.
4–44
CHAPTER 4 Polynomial and Rational Functions
y
y
40
32
24
16
8
54321
8
16
24
32
40
1 2 3 4 5 x
80
64
48
32
16
54321
16
32
48
64
80
d.
f.
1 2 3 4 5 x
1 2 3 4 5 x
y
40
32
24
16
8
54321
8
16
24
32
40
1 2 3 4 5 x
Sketch the graph of each function using the degree, endbehavior, x- and y-intercepts, zeroes of multiplicity, and
a few midinterval points to round-out the graph.
Connect all points with a smooth, continuous curve.
43. f 1x2 1x 32 1x 12 1x 22
44. g1x2 1x 22 1x 42 1x 12
45. p1x2 1x 12 2 1x 32
46. q1x2 1x 22 1x 22 2
47. Y1 1x 12 2 13x 22 1x 32
48. Y2 1x 22 1x 12 2 15x 22
49. r1x2 1x 12 2 1x 22 2 1x 12
50. s1x2 1x 32 1x 12 2 1x 12 2
51. f 1x2 12x 32 1x 12 3
52. g1x2 13x 42 1x 12 3
53. h1x2 1x 12 3 1x 32 1x 22
54. H1x2 1x 32 1x 12 2 1x 22 2
55. Y3 1x 12 3 1x 12 2 1x 22
56. Y4 1x 32 1x 12 3 1x 12 2
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Use the Guidelines for Graphing Polynomial Functions to
graph the polynomials.
57. y x3 3x2 4
58. y x3 13x 12
59. f 1x2 x3 3x2 6x 8
60. g1x2 x3 2x2 5x 6
61. h1x2 x3 x2 5x 3
62. H1x2 x3 x2 8x 12
63. p1x2 x4 10x2 9
64. q1x2 x 13x 36
4
2
65. r1x2 x4 9x2 4x 12
66. s1x2 x 5x 20x 16
4
3
67. Y1 x4 6x3 8x2 6x 9
68. Y2 x4 4x3 3x2 10x 8
69. Y3 3x4 2x3 36x2 24x 32
70. Y4 2x4 3x3 15x2 32x 12
71. F1x2 2x 3x 9x
4
3
2
72. G1x2 3x4 2x3 8x2
73. f 1x2 x5 4x4 16x2 16x
74. g1x2 x5 3x4 x3 3x2
䊳
351
Section 4.3 Graphing Polynomial Functions
75. h1x2 x6 2x5 4x4 8x3
76. H1x2 x6 3x5 4x4
In preparation for future course work, it becomes
helpful to recognize the most common square roots
in mathematics: 12 ⬇ 1.414, 13 ⬇ 1.732, and
16 ⬇ 2.449. Graph the following polynomials on a
graphing calculator, and use the calculator to locate the
maximum/minimum values and all zeroes. Use the zeroes
to write the polynomial in factored form, then verify the
y-intercept from the factored form and polynomial form.
77. h1x2 x5 4x4 9x 36
78. H1x2 x5 5x4 4x 20
79. f 1x2 2x5 5x4 10x3 25x2 12x 30
80. g1x2 3x5 2x4 24x3 16x2 36x 24
Use the graph of each function to construct its equation in
factored form and in polynomial form. Be sure to check
the y-intercept and adjust the lead coefficient if necessary.
81.
y
7
6
5
4
3
2
1
54321
1
2
3
1 2 3 4 5 x
82.
y
5
4
3
2
1
54321
1
2
3
4
5
1 2 3 4 5 x
WORKING WITH FORMULAS
83. Roots tests for a quartic polynomial
ax4 ⴙ bx3 ⴙ cx2 ⴙ dx ⴙ e:
b2 ⴚ 2ac
1r1 2 2 ⴙ 1r2 2 2 ⴙ 1r3 2 2 ⴙ 1r4 2 2 ⴝ
a2
In the Chapter 3 Reinforcing Basic Concepts
feature, we used relationships between the roots of
a quadratic equation and its coefficients to verify
the roots without having to substitute. Similar
root/coefficient relationships exist for cubic and
quartic polynomials, but the method soon becomes
too time consuming (see Exercise 94). There is
actually a little known formula for checking the
roots of a quartic polynomial (and others) that is
much more efficient. Given that r1, r2, r3, and r4
are the roots of the polynomial, the sum of the
b2 2ac
.
a2
Note that if a 1, the formula reduces to b2 2c.
(a) Use this test to verify that x 3, 1, 2, and 4
are the roots of x4 2x3 13x2 14x 24 0,
then (b) use these roots and the factored form to write
the equation in polynomial form to confirm results.
squares of the roots must be equal to
84. It is worth noting that the root test in Exercise 83
still applies when the roots are irrational and/or
complex. Use this test to verify that
x 13, 13, 1 2i, and 1 2i are the
solutions to x4 2x3 2x2 6x 15 0, then
use these zeroes and the factored form to write the
equation in polynomial form to confirm results.
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䊳
CHAPTER 4 Polynomial and Rational Functions
4–46
APPLICATIONS
85. Traffic volume: Between the hours of 6:00 A.M.
and 6.00 P.M., the volume of traffic at a busy
intersection can be modeled by the polynomial
v1t2 t4 25t3 192t2 432t, where v(t)
represents the number of vehicles above/below
average, and t is number of hours past 6:00 A.M.
(6:00 A.M. S 02. (a) Use the remainder theorem to
find the volume of traffic during rush hour
(8:00 A.M.), lunch time (12 noon), and the trip
home (5:00 P.M.). (b) Use the rational zeroes
theorem to find the times when the volume of
traffic is at its average 3 v1t2 04 . (c) Use this
information to graph v(t), then use the graph to
estimate the maximum and minimum flow of traffic
and the time at which each occurs.
86. Insect population: The population of a certain
insect varies dramatically with the weather, with
springlike temperatures causing a population
boom and extreme weather (summer heat and
winter cold) adversely affecting the population.
This phenomena can be modeled by the polynomial
p1m2 m4 26m3 217m2 588m, where p(m)
represents the number of live insects (in hundreds
of thousands) in month m 1m 僆 10, 1 4 S Jan2 .
(a) Use the remainder theorem to find the
population of insects during the cool of spring
(March) and the fair weather of fall (October).
(b) Use the rational zeroes theorem to find the
times when the population of insects becomes
dormant 3 p1m2 04 . (c) Use this information to
graph p(m), then use the graph to estimate the
maximum and minimum population of insects, and
the month at which each occurs.
87. Balance of payments: The graph shown represents
the balance of payments (surplus versus deficit) for
a large county over a 9-yr period. Use it to answer
the following:
Balance
a. What is the minimum
10 (10,000s)
8
possible degree
(9.5, ~6)
6
4
polynomial that can
2
Year
model this graph?
2 1 2 3 4 5 6 7 8 9 10
b. How many years did this 4
6
8
county run a deficit?
10
c. Construct an equation
model in factored form and in polynomial
form, adjusting the lead coefficient as needed.
How large was the deficit in year 8?
88. Water supply: The graph shown represents the
water level in a reservoir (above and below normal)
that supplies water to a metropolitan area, over a
6-month period. Use it to answer the following:
Level
a. What is the
10
(inches)
8
minimum possible
6
degree polynomial
4
2
Month
that can model this
1
2
3
4
5
6
2
graph?
4
6
b. How many months
8
10
was the water level
below normal in this
6-month period?
c. At the beginning of this period 1m 02, the
water level was 36 in. above normal, due to a
long period of rain. Use this fact to help
construct an equation model in factored form
and in polynomial form, adjusting the lead
coefficient as needed. Use the equation to
determine the water level in months three
and five.
89. In order to
Time
Volume
determine if the
(6:00
A.M. S 0)
(vehicles/min)
number of lanes
0
0
on a certain
highway should
2
222
be increased, the
4
100
flow of traffic (in
6
114
vehicles per min)
8
360
is carefully
10
550
monitored from
11
429
6:00 A.M. to
6:00 P.M. The
data collected are shown the table (6:00 A.M.
corresponds to t 0). (a) Draw a scatterplot and
decide on an appropriate form of regression, then
find the regression equation and graph the function
and scatterplot on the same screen. (b) Use the
regression equation and its graph to find the
maximum flow of traffic for the morning and
evening rush hours. (c) During what time(s) of day
is the flow rate 350 vehicles per hour?
90. The Goddard Memorial Rocket
Velocity
Club is testing a new two-stage Time
(sec)
(ft/sec)
rocket. Using a specialized
0
0
tracking device, the velocity of
1
441
the rocket is monitored every
second for the first 4.5 sec of
2
484
flight, with the data collected
3
459
in the table shown. (a) Draw a
4
696
scatterplot and decide on an
appropriate form of regression, then find the
regression equation and graph the function and
scatterplot on the same screen. (b) Use the
regression equation and its graph to find how many
seconds elapsed before the first stage burned out,
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Section 4.3 Graphing Polynomial Functions
and the rocket’s velocity at this time, then
(c) determine how many seconds elapsed (after
liftoff) until the second stage ignited. (d) At a
velocity of 1000 ft/sec, the fuel was exhausted and
the return chutes deployed. How many seconds
after liftoff did this occur?
91. A posh restaurant
Time
Customer
in a thriving
(10 A.M. S 0)
count
neighborhood opens
0
0
at 10 A.M. for the
2
79
lunch crowd, and
closes at 9 P.M. as the
4
41
dinner crowd leaves.
6
43
In order to ensure
9
122
that an adequate
11
3
number of cooks and
servers are available,
their hourly customer count is monitored each day
for 1 month with the data averaged and compiled in
the table shown. (a) Draw a scatterplot and decide
on an appropriate form of regression, then find the
regression equation and graph the function and
scatterplot on the same screen. (b) Use the
regression equation and its graph to find what time
the restaurant reaches its morning peak and its
evening peak. (c) At what time is business slowest,
and how many customers are in the restaurant at
that time? (d) Between what times is the restaurant
serving 100 customers or more?
䊳
353
92. Using the wind to
Wind velocity
Power
generate power is
(mph)
(W)
becoming more and
20
419
more prevalent. While
25
623
most people are aware
30
635
that a wind turbine
generates more power
35
593
with a stronger wind,
40
639
many are not aware that
the generators are built with a stall mechanism to
protect the generator, blades, and infrastructure in
very high winds. This affects the actual power output
as the generator operates near its threshold. The
power output [in watts (W)] of a certain generator is
shown in the table for wind velocity v in miles per
hour. (a) Draw a scatterplot and decide on an
appropriate form of regression, then find the
regression equation and graph the function and
scatterplot on the same screen. (b) Use the regression
equation and its graph to find the maximum safe
power output for this generator, and the wind speed
at which this occurs. (c) What is the power output in
a 23 mph wind? (d) If the manufacturer stipulates
that the turbine will experience automatic shutdown
when power output exceeds 900 W, what is the
greatest wind speed this turbine can tolerate?
EXTENDING THE CONCEPT
93. As discussed in this section, the study of endbehavior looks at what happens to the graph of a
function as 冟x冟 S q. Notice that as 冟x冟 S q, both 1x
and x12 approach zero. This fact can be used to
study the end-behavior of polynomial graphs.
a. For f 1x2 x3 x2 3x 6, factoring out x3
gives the expression
1
6
3
f 1x2 x3a1 2 3 b. What happens
x
x
x
to the value of the expression as x S q?
As x S q ?
b. Factor out x4 from g1x2 x4 3x3 4x2 5x 1. What happens to the value of the
expression as x S q? As x S q? How does
this affirm the end-behavior must be up/up?
94. If u, v, w, and z represent the roots of the quartic
polynomial ax4 bx3 cx2 dx e 0, then
the following relationships are true:
(a) u v w z b, (b) u1v z2 v1w z2 w1u z2 c, (c) u1vw wz2 v1uz wz2 d, and (d) u # v # w # z e. Use
these tests to verify that x 3, 1, 2, 4 are the
solutions to x4 2x3 13x2 14x 24 0,
then use these zeroes and the factored form to write
the equation in polynomial form to confirm results.
95. For what value of c will three of the four real roots
of x4 5x3 x2 21x c 0 be shared by the
polynomial x3 2x2 5x 6 0?
Show the following equations have no rational roots.
96. x5 x4 x3 x2 2x 3 0
97. x5 2x4 x3 2x2 3x 4 0
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䊳
4–48
CHAPTER 4 Polynomial and Rational Functions
MAINTAINING YOUR SKILLS
98. (3.6) Given f 1x2 x2 2x and g1x2 1x , find the
compositions h1x2 1 f ⴰ g2 1x2 and H1x2 1g ⴰ f 2 1x2, then state the domain of each.
99. (3.1) By direct substitution, verify that x 1 2i
is a solution to x2 2x 5 0 and name the
second solution.
100. (Appendix A.3/Appendix A.6) Solve each of the
following equations.
a. 12x 52 16 x2 3 x 31x 22
b. 1x 1 3 12x 2
21
2
4
5 2
c.
x3
x 9
101. (1.3) Determine if the relation shown is a function.
If not, explain how the definition of a function is
violated.
feline
canine
dromedary
equestrian
bovine
cow
horse
cat
camel
dog
MID-CHAPTER CHECK
1. Compute 1x3 8x2 7x 142 1x 22 using
long division and write the result in two ways:
(a) dividend 1quotient2 1divisor2 remainder and
remainder
dividend
(b)
.
1quotient2 divisor
divisor
2. Given that x 2 is a factor of f 1x2 2x4 x3 8x2 x 6, use the rational zeroes theorem to
write f (x) in completely factored form.
3. Use the remainder theorem to evaluate f 122, given
f 1x2 3x4 7x2 8x 11.
4. Use the factor theorem to find a third-degree
polynomial having x 2 and x 1 i as roots.
5. Use the intermediate value theorem to show that
g1x2 x3 6x 4 has a root in the interval (2, 3).
6. Use the rational zeroes theorem, tests for 1 and 1,
synthetic division, and the remainder theorem to
write f 1x2 x4 5x3 20x 16 in completely
factored form.
7. Find all the zeroes of h, real and complex:
h1x2 x4 3x3 10x2 6x 20.
8. Sketch the graph of p using its degree, end-behavior,
y-intercept, zeroes of multiplicity, and any
midinterval points needed, given
p1x2 1x 12 2 1x 12 1x 32.
9. Use the Guidelines for Graphing to draw the graph
of q1x2 x3 5x2 2x 8.
10. When fighter pilots train for dogfighting, a “harddeck” is usually established below which no
competitive activity can take place. The polynomial
graph given shows Maverick’s altitude above and
below this hard-deck during a 5-sec interval.
Altitude
a. What is the minimum
A (100s of feet)
15
possible degree
12
9
polynomial that could
6
form this graph? Why?
3
Seconds
b. How many seconds
(total) was Maverick
below the hard-deck for
these 5 sec of the
exercise?
3
6
9
12
15
1 2 3 4 5 6 7 8 9 10 t
c. At the beginning of this time interval (t 0),
Maverick’s altitude was 1500 ft above the harddeck. Use this fact and the graph given to help
construct an equation model in factored form and
in polynomial form, adjusting the lead coefficient
if needed. Use the equation to determine
Maverick’s altitude in relation to the hard-deck at
t 2 and t 4.
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Reinforcing Basic Concepts
355
REINFORCING BASIC CONCEPTS
Approximating Real Zer oes
Consider the equation x4 x3 x 6 0. Using the rational zeroes theorem, the possible rational zeroes are
51, 6, 2, 36. The tests for 1 and 1 indicate that neither is a zero: f 112 3 and f 112 7. Descartes’
rule of signs reveals there must be one positive real zero since the coefficients of f 1x2 change sign one time:
f 1x2 x4 x3 x 6, and one negative real zero since f 1x2 also changes sign one time: f 1x2 x4 x3 x 6. The remaining two zeroes must be complex. Using x 2 with synthetic division shows 2 is not a
zero, but the coefficients in the quotient row are all positive, so 2 is an upper bound:
2|
1
1
1
2
3
0
6
6
1
12
13
6
26
20
|
coefficients of f (x)
q (x)
Using x 2 shows that 2 is a zero and a lower bound for all other zeroes (quotient row alternates in sign):
2|
1
1
1
2
1
0
2
2
1
4
3
6
6
|0
coefficients of f (x)
q1(x)
This means the remaining real zero must be a positive irrational
number less than 2 (all other possible rational zeroes were eliminated). The
x
f(x)
Conclusion
quotient polynomial q1 1x2 x3 x2 2x 3 is not factorable, yet we’re
1
3
— Zero is here,
left with the challenge of finding this final zero. While there are many
1.5
3.94
use x 1.25
advanced techniques available for approximating irrational zeroes, at this
next
2
20
level either technology or a technique called bisection is commonly used.
The bisection method combines the intermediate value theorem with
x
f(x)
Conclusion
successively smaller intervals of the input variable, to narrow down the
3
1
location of the irrational zero. Although “bisection” implies halving the
Zero is here,
1.25 0.36
interval each time, any number within the interval will do. The bisection
— use x 1.30
1.5
3.94
method may be most efficient using a succession of short input/output
next
tables as shown, with the number of tables increased if greater accuracy is
x
f(x)
Conclusion
desired. Since f 112 3 and f 122 20, the intermediate value theorem
tells us the zero must be in the interval [1, 2]. We begin our search here,
1.25 0.36
— Zero is here,
rounding noninteger outputs to the nearest 100th. As a visual aid, positive
1.30
0.35
use x 1.275
outputs are in blue, negative outputs in red.
1.5
3.94
next
A reasonable estimate for the zero appears to be x 1.275. Evaluating
the function at this point gives f 11.2752 ⬇ 0.0097, which is very close
to zero. Naturally, a closer approximation is obtained using the capabilities of a graphing calculator. To seven decimal
places the zero is x ⬇ 1.2756822.
Exercise 1: Use the intermediate value theorem to show that f 1x2 x3 3x 1 has a zero in the interval [1, 2], then
use bisection to locate the zero to three decimal place accuracy.
Exercise 2: The function f 1x2 x4 3x 15 has two real zeroes in the interval 3 5, 54 . Use the intermediate value
theorem to locate the zeroes, then use bisection to find the zeroes accurate to three decimal places.
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Precalculus—
4.4
Graphing Rational Functions
LEARNING OBJECTIVES
In Section 4.4 you will see
how we can:
A. Locate the vertical
B.
C.
D.
E.
asymptotes and find the
domain of a rational
function
Apply the concept of
“roots of multiplicity” to
rational functions and
graphs
Find horizontal
asymptotes of rational
functions
Graph general rational
functions
Solve applications of
rational functions
Our first exposure to rational functions occurred in Section 2.4, where we looked at the
1
1
attributes and properties of the basic rational functions f 1x2 and g1x2 2 shown
x
x
in Figures 4.32 and 4.33.
Figure 4.33
1
g1x2 2
x
Figure 4.32
1
f1x2 x
y
y
5
5
4
4
3
3
2
2
1
1
5 4 3 2 1
1
1
2
3
4
5
x
5 4 3 2 1
1
2
2
3
3
4
4
5
5
1
2
3
4
5
x
Much of what we learned about these functions can be generalized and applied to the
general rational functions that follow. For convenience and emphasis, the definition of
a rational function is repeated here.
Rational Functions
A rational function V(x) is one of the form
V1x2 p1x2
d1x2
,
where p and d are polynomials and d1x2 0.
The domain of V(x) is all real numbers, except the zeroes of d.
Our study begins by taking a closer look at the zeroes of d(x) that are excluded from the
domain, and what happens to the graph of a rational function at or near these zeroes.
These observations will form a key component of graphing general rational functions.
A. Rational Functions and Vertical Asymptotes
The graphs shown in Figures 4.34 through 4.37 illustrate that rational graphs come in
many shapes, often in “pieces,” and exhibit asymptotic behavior.
WORTHY OF NOTE
In Section 2.5, we studied special
p1x2
cases of d1x2 , where p and d shared
a common factor, creating a “hole”
in the graph. Rational functions of
this form will be investigated further
in Section 4.5. In this section, we’ll
assume the functions are given in
simplest form (the numerator and
denominator have no common
factors).
356
Figure 4.34
2x
g1x2 2
x 1
Figure 4.35
3
w1x2 2
x 1
y
y
5
5
4
4
3
3
2
2
1
5 4 3 2 1
1
1
1
2
3
4
5
x
5 4 3 2 1
1
2
2
3
3
4
4
5
5
1
2
3
4
5
x
4–50
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Section 4.4 Graphing Rational Functions
Figure 4.36
x2
h1x2 2
x 2x 3
Figure 4.37
v 1x2 x2 4
x1
y
y
10
10
8
8
6
6
4
4
2
2
108 6 4 2
2
108 6 4 2
2
2
4
6
8 10
x
4
4
6
6
8
8
10
10
2
4
6
8 10
x
For the functions f 1x2 1x and g1x2 x12, a vertical asymptote occurred at the zero of
each denominator. This actually applies to all rational functions in simplified form. For
V1x2 p1x2
d1x2 , if c is a zero of d(x), the function can be evaluated at every point near c,
but not at c. This creates a break or discontinuity in the graph of V resulting in the asymptotic behavior.
Vertical Asymptotes of a Rational Function
Given V1x2 p1x2
is a rational function in simplest form,
d1x2
vertical asymptotes will occur at the real zeroes of d.
Breaks created by vertical asymptotes are said to be nonremovable, because
there is no way to repair the break, even if a piecewise-defined function were used. See
Example 5, Section 2.6.
EXAMPLE 1
䊳
Finding Vertical Asymptotes and the Domain of a Rational Function
Locate the vertical asymptote(s) of each function given, then state its domain.
2x
3
a. g1x2 2
b. w1x2 2
x 1
x 1
2
x
x2 4
c. h1x2 2
d. v1x2 x1
x 2x 3
Solution
䊳
a. Setting the denominator equal to zero gives x2 1 0, so vertical
asymptotes will occur at x 1 and x 1. The domain of g is
x 僆 1q, 12 ´ 11, 12 ´ 11, q 2 . See Figure 4.34.
b. Since the equation x2 1 0 has no real zeroes, there are no vertical
asymptotes and the domain of w is unrestricted: x 僆 R. See Figure 4.35.
c. Solving x2 2x 3 0 gives 1x 12 1x 32 0, with solutions x 1
and x 3. There are vertical asymptotes at x 1 and x 3, and the
domain of h is x 僆 1q, 12 ´ 11, 32 ´ 13, q 2 . See Figure 4.36.
d. Solving x 1 0 gives x 1, and a vertical asymptote will occur at
x 1. The domain of v is x 僆 1q, 12 ´ 11, q 2 . See Figure 4.37.
Now try Exercises 7 through 14
䊳
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4–52
CHAPTER 4 Polynomial and Rational Functions
Using a TABLE for the function g(x) from Example 1(a), we again note that we’re able
to evaluate rational functions for all values near the zeroes of the denominator, but not
at these zeroes. See Figures 4.38 through 4.40.
Figure 4.38
Figure 4.39
A. You’ve just seen how we
can locate vertical asymptotes
and find the domain of a
rational function
B. Vertical Asymptotes and Multiplicities
The “cross” and “bounce” concept used for polynomial graphs can also be applied to
rational graphs, particularly when viewed in terms of sign changes in the dependent
1
variable. As you can see in Figures 4.41 to 4.43, the function f 1x2 changes
x2
sign at the asymptote x 2 (negative on one side, positive on the other), and the zero
1
of the denominator has multiplicity 1 (odd). The function g1x2 does not
1x 12 2
change sign at the asymptote x 1 (positive on both sides), and its denominator has multiplicity 2 (even). As with our earlier study of multiplicities, when these two are combined
1
into the single function v1x2 , the function still changes sign at
1x 22 1x 12 2
x 2, and does not change sign at x 1.
Figure 4.42
1
g1x2 1x 12 2
Figure 4.41
1
f 1x2 x2
Figure 4.43
1
h1x2 1x 221x 12 2
y
y
y
5
5
5
4
4
4
3
3
3
2
2
1
1
7 6 5 4 3 2 1
1
f(x)
Figure 4.40
1
2
3
x
5 4 3 2 1
1
1
2
3
1
4
5
x
5 4 3 2 1
1
2
2
3
3
4
4
4
5
5
5
EXAMPLE 2
䊳
h(x) 0
2
g(x)
h(x) 0
1
2
3
4
5
x
2
3
Finding Sign Changes at Vertical Asymptotes
Locate the vertical asymptotes of each function and state whether the function will
change sign from one side of the asymptote(s) to the other.
x2 4x 4
x2 2
a. f 1x2 2
b. g1x2 2
x 2x 3
x 2x 1
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359
Section 4.4 Graphing Rational Functions
Solution
䊳
a. Factoring the denominator of f and setting it equal to zero gives
1x 12 1x 32 0, and vertical asymptotes will occur at x 1 and x 3
(both multiplicity 1). The function will change sign at each asymptote (see
Figure 4.44). Factoring the numerator gives 1x 22 2 0, and the graph will
“bounce” off the x-axis at the zero x 2 (multiplicity even — no sign change).
b. Factoring the denominator of g and setting it equal to zero gives 1x 12 2 0.
There will be a vertical asymptote at x 1, but the function will not change
sign since it’s a zero of even multiplicity (see Figure 4.45).
Figure 4.45
x2 2
g1x2 2
x 2x 1
Figure 4.44
x2 4x 4
f1x2 2
x 2x 3
y
y
10
8
8
7
6
6
4
108 6 4 2
2
5
f(x)
2
2
4
6
8 10
x
4
6
B. You’ve just seen how
we can apply the concept of
“roots of multiplicity” to
rational functions and graphs
g(x)
4
3
2
1
8
7 6 5 4 3 2 1
1
10
2
1
2
3
x
Now try Exercises 15 through 20
䊳
C. Finding Horizontal Asymptotes
A study of horizontal asymptotes is closely related to our study of “dominant terms” in
Section 4.3. Recall the highest degree term in a polynomial tends to dominate all other
2x2 4x 3
terms as 冟x冟 S q . For v1x2 2
, both polynomials have the same degree,
x 2x 1
2x2
2x2 4x 3
⬇ 2 2 for large values of x: as 冟x冟 S q, y S 2 and y 2 is a
so 2
x 2x 1
x
horizontal asymptote for v. When the degree of the numerator is smaller than the
degree of the denominator, our earlier work with y 1x and y x12 showed there was a
horizontal asymptote at y 0 (the x-axis), since as 冟x冟 S q, y S 0. In general,
LOOKING AHEAD
In Section 4.5 we will explore
two additional kinds of
asymptotic behavior,
(1) oblique (slant) asymptotes
and (2) asymptotes that are
nonlinear.
Horizontal Asymptotes
Given V1x2 p1x2
is a rational function in lowest terms, where the leading term
d1x2
of p is axn and the leading term of d is bxm (p has degree n, d has degree m).
I. If n 6 m, there is a horizontal asymptote at y 0 (the x-axis).
II. If n m, there is a horizontal asymptote at y ab .
III. If n 7 m, the graph has no horizontal asymptote.
Finally, while the graph of a rational function can never “cross” the vertical
asymptote x h (since the function simply cannot be evaluated at h), it is possible for
a graph to cross the horizontal asymptote y k (some do, others do not). To find out
which is the case, we set the function equal to k and solve.
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CHAPTER 4 Polynomial and Rational Functions
EXAMPLE 3
Locating Horizontal Asymptotes
䊳
Locate the horizontal asymptote for each function, if one exists. Then determine if
the graph will cross the asymptote.
x2 4
3x
3x2 x 6
a. f 1x2 2
b. g1x2 2
c. v1x2 2
x 2
x 1
x x6
Solution
䊳
a. For f (x), the degree of the numerator 6 degree of the denominator, indicating
a horizontal asymptote at y 0. Solving f 1x2 0, we find x 0 is the only
solution and the graph will cross the horizontal asymptote at (0, 0) (see
Figure 4.46).
b. For g(x), the degree of the numerator and the denominator are equal. This
x2
means g1x2 ⬇ 2 1 for large values of x, and there is a horizontal asymptote
x
at y 1. Solving g1x2 1 gives
x2 4
1
x2 1
x2 4 x2 1
4 1
y 1 S horizontal asymptote
multiply by x2 1
no solution
The graph will not cross the asymptote (see Figure 4.47).
c. For v(x), the degree of the numerator and denominator are once again equal,
3x2
so v1x2 ⬇ 2 3 and there is a horizontal asymptote at y 3.
x
Solving v1x2 3 gives
3x2 x 6
3
x2 x 6
3x2 x 6 31x2 x 62
3x2 x 6 3x2 3x 18
4x 12 0
x3
y 3 S horizontal asymptote
multiply by x2 x 6
distribute
simplify
result
The graph will cross its asymptote at x 3 (see Figure 4.48).
Figure 4.46
3x
f 1x2 2
x 2
Figure 4.48
3x 2 x 6
v1x2 2
x x6
Figure 4.47
x2 4
g1x2 2
x 1
y
y
y
10
10
10
8
8
8
6
6
6
4
4
4
2
2
2
108 6 4 2
2
2
4
6
8 10
x
5 4 3 2 1
2
1
2
3
4
5
x
108 6 4 2
2
4
4
4
6
6
6
8
8
8
10
10
10
2
4
6
8 10
x
Now try Exercises 21 through 26
䊳
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Section 4.4 Graphing Rational Functions
Using the TABLE feature once again gives a
numerical confirmation of asymptotic behavior in
the horizontal direction. For f (x) from Example 3,
the table verifies that larger positive values of x
produce smaller and smaller values of f (x): as
x S q, f 1x2 S 0 (Figure 4.49). For v(x), a table will
confirm that the graph passes through the horizontal
asymptote at the point (3, 3) (Figure 4.50), but still approaches the asymptote y 3 as x becomes very
large: as x S q , v1x2 S 3 (Figure 4.51).
Figure 4.50
C. You’ve just seen how
we can find the horizontal
asymptotes of rational
functions
361
Figure 4.49
Figure 4.51
Finally, it’s helpful to note that the location of all nonvertical asymptotes and whether
or not the graph crosses through them can actually be found using division. The quotient q(x) gives the equation of the asymptote, and the zeroes of the remainder r(x) will
indicate if and where the graph and asymptote will cross. From Example 3(c), long
division gives q1x2 3 and r1x2 4x 12 (verify this), showing there is a horizontal asymptote at y 3, which the graph crosses at x 3 since r 132 0. See Exercises 27 through 30.
D. The Graph of a Rational Function
Our observations to this point lead us to this general strategy for graphing rational
functions. Not all graphs require every step, but together they provide an effective
approach.
Guidelines for Graphing Rational Functions
3d1x2 0 4 is a rational function in lowest terms,
d1x2
1. Find V(0) (if it exists):
2. Find the zeroes of p (if they exist):
the y-intercept
the x-intercept(s).
3. Find the zeroes of d (if they exist):
4. Locate the horizontal asymptote
the vertical asymptotes.
if it exists.
5. Determine if the graph will cross
6. Compute any additional points
the horizontal asymptote.
needed to sketch the graph.
Given V1x2 p1x2
• Draw the asymptotes, plot the intercepts and additional points, and determine
where V(x) changes sign to complete the graph.
As you work to complete the graph, it helps to keep the following in mind:
• The graph must go through its plotted points and approach the asymptotes.
• The graph may cross a horizontal asymptote, but never a vertical asymptote.
• Function values must change sign at any zero of odd multiplicity.
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CHAPTER 4 Polynomial and Rational Functions
EXAMPLE 4
䊳
Graphing Rational Functions
Graph each function given.
x2 x 6
a. f 1x2 2
x x6
Solution
䊳
2x2 4x 2
x2 7
1x 22 1x 32
a. Begin by writing f in factored form: f 1x2 .
1x 32 1x 22
1. y-intercept: f 102 2.
3.
4.
5.
b. g1x2 122 132
1, so the y-intercept is (0, 1).
132 122
x-intercepts: Setting the numerator equal to zero gives 1x 221x 32 0,
showing the x-intercepts will be (2, 0) and (3, 0).
Vertical asymptote(s): Setting the denominator equal to zero gives
1x 32 1x 22 0, showing there will be vertical asymptotes at
x 3 and x 2.
Horizontal asymptote: Since the degree of the numerator and the degree
x2
of the denominator are equal, y 2 1 is a horizontal asymptote.
x
x2 x 6
1
Solving 2
f 1x2 1 S horizontal asymptote
x x6
x2 x 6 x2 x 6 multiply by x2 x 6
simplify
2x 0
x0
solve
The graph will cross the horizontal asymptote at (0, 1).
The information from steps 1 through 5 is shown in Figure 4.52, and
indicates we have no information about the graph in the interval (q, 3).
Since rational functions are defined for all real numbers except the zeroes of
d, we know there must be a “piece” of the graph in this interval.
6. Selecting x 4 to compute one additional point, we find
122 172
7
7
f 142 112 162 14
6 3 . The point is (4, 3 ).
All factors of f are linear, so function values will alternate sign in the intervals
created by x-intercepts and vertical asymptotes. The y-intercept (0, 1) shows
f (x) is positive in the interval containing 0. To meet all necessary conditions,
we complete the graph as shown in Figure 4.53.
Figure 4.52
Figure 4.53
y
y x2
x 3
5
5
冢4, g冣
y1
pos
n
e
g
pos
5
n
e
g
pos
x
5
5
5
x
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Precalculus—
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b. Writing g in factored form gives g1x2 2.
3.
4.
5.
It’s useful to note that the number
of “pieces” forming a rational graph
will always be one more than the
number of vertical asymptotes. The
3x
graph of f1x2 2
(Figure 4.46)
x 2
has no vertical asymptotes and one
piece, y 1x has one vertical
asymptote and two pieces,
2
4
g1x2 x 2
(Figure 4.47) has two
x 1
vertical asymptotes and three
pieces, and so on.
21x2 2x 12
x2 7
2112 2
21x 12 2
1x 1721x 172
.
2
2
. The y-intercept is a 0, b.
7
7
1 172 1 172
x-intercept(s): Setting the numerator equal to zero gives 21x 12 2 0,
with x 1 as a zero of multiplicity 2. The x-intercept is (1, 0).
Vertical asymptote(s): Setting the denominator equal to zero gives
1x 172 1x 172 0, showing there will be asymptotes at x 17
and x 17.
Horizontal asymptote: The degree of the numerator is equal to the degree
2x2
of denominator, so y 2 2 is a horizontal asymptote.
x
2
2x 4x 2
2
Solve
g 1x2 2 S horizontal asymptote
x2 7
2x2 4x 2 2x2 14 multiply by x 2 7
simplify
4x 16
x4
solve
1. y-intercept: g102 WORTHY OF NOTE
363
Section 4.4 Graphing Rational Functions
The graph will cross its horizontal
asymptote at (4, 2). The information
from steps 1 to 5 is shown in Figure 4.54,
and indicates we have no information
about the graph in the interval
(q, 17).
215 12 2
6. Selecting x 5, g152 152 2 7
2162 2
25 7
21362
18
4
Figure 4.54
y x 兹7
x 兹7
5
(4, 2)
y2
neg
6
(1, 0)
pos
pos
6
x
5
The point (5, 4) is on the graph
(Figure 4.55).
Figure 4.55
y
Since factors of the denominator have odd
multiplicity, function values will alternate sign
on either side of the asymptotes. The factor in
the numerator has even multiplicity, so the
graph will “bounce off” the x-axis at x 1
(no change in sign). The y-intercept (0, 27 )
shows the function is negative in the interval
containing 0. This information and the
completed graph are shown in Figure 4.55.
(5, 4)
5
6
5
x
5
Now try Exercises 31 through 54
䊳
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CHAPTER 4 Polynomial and Rational Functions
Examples 3 and 4 demonstrate that graphs of rational functions come in a large variety. Once the components of the graph have been found, completing the graph presents
an intriguing and puzzle-like challenge as we attempt to sketch a graph that meets all
conditions. As we’ve done with other functions, can you reverse this process? That is,
given the graph of a rational function, can you construct its equation?
EXAMPLE 5
䊳
Finding the Equation of a Rational Function from Its Graph
Use the graph of f (x) shown to construct its equation.
Solution
䊳
The x-intercepts are (1, 0) and (4, 0), so the
numerator must contain the factors (x 1) and
(x 4). The vertical asymptotes are x 2
and x 3, so the denominator must have the
factors 1x 22 and 1x 32 . So far we have:
f 1x2 y
5
a1x 12 1x 42
(2, 3)
1x 22 1x 32
5
5
Since (2, 3) is on the graph, we substitute 2 for
x and 3 for f (x) to solve for a:
a12 12 12 42
substitute 3 for f (x) and 2 for x
3
12 22 12 32
5
3a
3
simplify
2
2a
solve
21x 12 1x 42
2x2 6x 8
2
The result is f 1x2 , with a horizontal
1x 22 1x 32
x x6
asymptote at y 2 and a y-intercept of (0, 43 ), which fit the graph very well.
Now try Exercises 55 through 58
x
䊳
As a final note, there are many rational graphs that have x- and y-intercepts and vertical asymptotes, but no horizontal asymptotes. Two examples are shown in Figures 4.56
and 4.57.
Figure 4.56
Figure 4.57
2
x 2 14
x 4
f1x2 g1x2 x1
x1
f (x)
12
8
10
6
8
4
6
2
4
108 6 4 2
2
D. You’ve just seen how
we can graph general rational
functions
g(x)
10
2
2
4
6
8 10
x
4
4 3 2 1
2
6
4
8
6
10
8
1
2
3
4
5
6
x
Without more information regarding nonhorizontal asymptotes, sketching these graphs
requires a substantial number of plotted points. Rational graphs of this type will be
studied in more detail in Section 4.5, enabling us to complete each graph using far
fewer points. See Exercises 59 through 62.
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College Algebra Graphs & Models—
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Section 4.4 Graphing Rational Functions
365
E. Applications of Rational Functions
In many applications of rational functions, the coefficients can be rather large and the
graph should be scaled appropriately.
EXAMPLE 6
䊳
Modeling the Rides at an Amusement P ark
A popular amusement park wants to add new rides and asks various contractors to
submit ideas. Suppose one ride engineer offers plans for a ride that begins with a
near vertical drop into a dark tunnel, quickly turns and becomes more horizontal,
pops out the tunnel’s end, then coasts up to the exit platform, braking 20 m from
the release point. The height of a rider above ground is modeled by the function
39x2 507x 468
h1x2 , where h(x) is the height in meters at a horizontal
3x2 23x 20
distance of x meters from the release point.
a. Graph the function for x 僆 31, 20 4 .
b. How high is the release point for this ride?
c. How long is the tunnel from entrance to exit?
d. What is the height of the exit platform?
Solution
䊳
468
or 23.4 m. Also, since
20
the degree of the numerator is equal to that of the denominator, the ratio of
39x2
Figure 4.58
leading terms 2 indicates a horizontal
h(x)
3x
32
asymptote at y 13. Writing h(x) in
391x 12 1x 122
24
factored form gives h1x2 ,
13x 202 1x 12
16
showing the x-intercepts will be (1, 0) and
(12, 0), with vertical asymptotes at x 6.6
8
and x 1. Computing midinterval points of
x 4, 8, and 18 gives (4, 5.85), (8, 2.76),
4
8
12
16
20
and (18, 2.83). Graphing the function over the
8
specified interval produces the graph shown
in Figure 4.58.
Figure 4.59
b. From the context, the release point is at 23.4 m.
c. The ride enters the tunnel at x 1 and exits
at x 12, making the tunnel 11 m long.
d. Since the ride begins braking at a distance of
20 m, the platform must be h1202 ⬇ 3.5 m
high. See Figure 4.59.
a. Here we begin by noting the y-intercept is h102 Now try Exercises 65 through 76
E. You’ve just seen how
we can solve applications of
rational functions
x
䊳
As a final note, the ride proposed in Example 6 was never approved due to excessive
g-forces on the riders. Example 6 helps to illustrate that when it comes to applications
of rational functions, portions of the graph may be ignored due to the context. In addition, some applications may focus on a specific attribute of the graph, such as the horizontal asymptotes in Exercises 65, 66, and elsewhere, or the vertical asymptotes in
Exercises 67 and 68.
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CHAPTER 4 Polynomial and Rational Functions
4.4 EXERCISES
䊳
CONCEPTS AND VOCABULAR Y
Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.
1. Write the following in direction/approach notation.
As x becomes an infinitely large negative number, y
approaches 2.
2. For any constant k, the notation as 冟x冟 S q,
y S k is an indication of a
asymptote,
while x S k, 冟y冟 S q indicates a
asymptote.
3. Vertical asymptotes are found by setting the
equal to zero. The x-intercepts are found
by setting the
equal to zero.
4. If the degree of the numerator is equal to the
degree of the denominator, a horizontal asymptote
occurs at y ab , where ab represents the ratio of
the
.
3x2 2x
and a table of
2x2 3
values to discuss the concept of horizontal
asymptotes. At what positive value of x is the graph
of g within 0.01 of its horizontal asymptote?
5. Use the function g1x2 䊳
6. Name all of the “tools” at your disposal that play a
role in the graphing of rational functions. Which
tools are indispensable and always used? Which are
used only as the situation merits?
DEVELOPING YOUR SKILLS
Give the location of the vertical asymptote(s) if they
exist, and state the function’s domain.
7. f 1x2 x2
x3
9. g1x2 3x2
x2 9
8. F1x2 4x
2x 3
10. G1x2 x1
9x2 4
11. h1x2 x2 1
x5
12. H1x2 2
2x 3x 5
2x x 3
13. p1x2 2x 3
2
x x1
2
14. q1x2 2x3
x2 4
Give the location of the vertical asymptote(s) if they
exist, and state whether function values will change sign
(positive to negative or negative to positive) from one
side of the asymptote to the other.
15. Y1 x1
x x6
2
17. r1x2 16. Y2 2x 3
x x 20
2
x2 3x 10
x2 2x 15
18.
R1x2
x2 6x 9
x2 4x 4
19. Y1 x
x 2x 4x 8
20. Y2 2x
x3 x2 x 1
3
2
For the functions given, (a) determine if a horizontal
asymptote exists and (b) determine if the graph will
cross the asymptote, and if so, where it crosses.
21. Y1 2x 3
x2 1
22. Y2 4x 3
2x2 5
23. r1x2 4x2 9
2x2 x 10
24.
R1x2
x2 3x 18
x2 5
25. p1x2 3x2 5
x2 1
26. P1x2 3x2 5x 2
x2 4
Apply long division to find the quotient and remainder
for each function. Use this information to determine the
equation of the horizontal asymptote, and whether the
graph will cross this asymptote. Verify answers by
graphing the functions on a graphing calculator and
locating points of intersection.
27. v1x2 8x
x 1
28. f 1x2 4x 8
x2 1
29. g1x2 2x2 8x
x2 4
30. h1x2 x2 x 6
x2 1
2
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Give the location of the x- and y-intercepts (if they
exist), and discuss the behavior of the function (bounce
or cross) at each x-intercept.
x2 3x
31. f 1x2 2
x 5
2x x2
32. F1x2 2
x 2x 3
33. g1x2 x2 3x 4
x2 7x 6
34.
G1x2
x2 1
x2 2
35. h1x2 x3 6x2 9x
4x 4x2 x3
36. H1x2 2
4x
x2 1
Use the Guidelines for Graphing Rational Functions to
graph the functions given.
x3
37. f 1x2 x1
39. F1x2 8x
x 4
2
2x
41. p1x2 2
x 4
2
40. G1x2 12x
x2 3
3x
2x
46. H1x2 2
x2 6x 9
x 2x 1
x1
x 3x 4
49. s1x2 4x2
2x2 4
x2 4
51. Y1 2
x 1
54. V1x2 3x
x x x1
48. Y2 1x
x2 2x
50. S1x2 2x2
x2 1
x2 x 12
52. Y2 2
x x 12
3
3
2
Use the vertical asymptotes, x-intercepts, and their
multiplicities to construct an equation that corresponds
to each graph. Be sure the y-intercept estimated from
the graph matches the value given by your equation for
x ⴝ 0. Check work on a graphing calculator.
55.
56.
y
10
8
6
4
2
108 642
2
57.
2 4 6 8 10 x
4
6
8
10
58.
y
4
6
8
10
10
8
6
4
2
108642
2
2 4 6 8 10 x
10
8
6
4
2
108642
2
y
(2, 2)
(1, 1)
4
6
8
10
3x
42. P1x2 2
x 9
45. h1x2 2
2x
x 2x2 4x 8
2
2x x2
x2 3x
44.
Q1x2
x2 4x 5
x2 2x 3
47. Y1 53. v1x2 x4
38. g1x2 x2
43. q1x2 䊳
367
Section 4.4 Graphing Rational Functions
4 6 8 10 x
冢5, Ò
16 冣
y
10
8
6
4
2
108642
2
(3, 4)
2 4 6 8 10 x
4
6
8
10
Graph the following using the Guidelines for Graphing
Rational Functions. From the equations given, note
there are no horizontal asymptotes (the degree of each
numerator is greater than the degree of the
denominator) and a large number of plotted points may
be necessary to complete each graph.
59. v1x2 x2 4
x
60. f 1x2 9 x2
x1
61. g1x2 x2
x1
62. h1x2 1 x2
x2
WORKING WITH FORMULAS
ax
x ⴙb
The population density of urban areas (in people
per square mile) can be modeled by the formula
shown, where a and b are constants related to the
overall population and sprawl of the area under
study, and D(x) is the population density (in
hundreds), x mi from the center of downtown.
63. Population density: D1x2 ⴝ
2
Graph the function for a 63 and b 20 over the
interval x 僆 3 0, 504 , and then use the graph to
answer the following questions.
a. What is the significance of the horizontal
asymptote (what does it mean in this context)?
b. How far from downtown does the population
density fall below 525 people per square mile?
How far until the density falls below 300
people per square mile?
c. Use the graph and a table to determine how far
from downtown the population density reaches
a maximum. What is this maximum?
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CHAPTER 4 Polynomial and Rational Functions
kx
100 ⴚ x
Some industries resist cleaner air standards because
the cost of removing pollutants rises dramatically as
higher standards are set. This phenomenon can be
modeled by the formula given, where C(x) is the
cost (in thousands of dollars) of removing x% of the
pollutant and k is a constant that depends on the type
of pollutant and other factors.
Graph the function for k 250 over the interval
x 僆 3 0, 1004 , and then use the graph to answer the
following questions.
4–62
64. Cost of removing pollutants: C1x2 ⴝ
䊳
a. What is the significance of the vertical
asymptote (what does it mean in this context)?
b. If new laws are passed that require 80% of a
pollutant to be removed, while the existing law
requires only 75%, how much will the new
legislation cost the company? Compare the
cost of the 5% increase from 75% to 80% with
the cost of the 1% increase from 90% to 91%.
c. What percent of the pollutants can be removed
if the company budgets 2250 thousand dollars?
APPLICATIONS
C(h)
65. Medication in the blood0.4
stream: The concentration C 0.3
0.2
of a certain medicine in the
0.1
bloodstream h hours after
2 6 10 14 18 22 26 h
being injected into the
shoulder is given by the
2h2 h
. Use the given graph of
function: C1h2 3
h 70
the function to answer the following questions.
a. Approximately how many hours after injection
did the maximum concentration occur? What
was the maximum concentration?
b. Use C(h) to compute the rate of change for the
intervals h 8 to h 10 and h 20 to
h 22. What do you notice?
c. Use mathematical notation to state what
happens to the concentration C as the number of
hours becomes infinitely large. What role does
the h-axis play for this function?
66. Supply and demand: In
S(t)
10.0
response to certain market
7.5
demands, manufacturers will 5.0
2.5
quickly get a product out on
the market to take advantage
10 20 30 40 50 60 70 t
of consumer interest. Once
the product is released, it is not uncommon for
sales to initially skyrocket, taper off, and then
gradually decrease as consumer interest wanes. For
a certain product, sales can be modeled by the
250t
, where S(t) represents the
function S1t2 2
t 150
daily sales (in $10,000) t days after the product has
debuted. Use the given graph of the function to
answer the following questions.
a. Approximately how many days after the
product came out did sales reach a maximum?
What was the maximum sales?
b. Use S(t) to compute the rate of change for the
intervals t 7 to t 8 and t 60 to t 62.
What do you notice?
c. Use mathematical notation to state what
happens to the daily sales S as the number of
days becomes infinitely large. What role does
the t-axis play for this function?
67. Cost to remove pollutants: For a certain coalburning power plant, the cost to remove pollutants
from plant emissions can be modeled by
80p
C1p2 , where C(p) represents the cost
100 p
(in thousands of dollars) to remove p percent of the
pollutants. (a) Find the cost to remove 20%, 50%,
and 80% of the pollutants, then comment on the
results; (b) graph the function using an appropriate
scale; and (c) use mathematical notation to state
what happens if the power company attempts to
remove 100% of the pollutants.
68. Costs of recycling: A large city has initiated a new
recycling effort, and wants to distribute recycling bins
for use in separating various recyclable materials.
City planners anticipate the cost of the program can
220p
, where
be modeled by the function C1p2 100 p
C(p) represents the cost (in $10,000) to distribute
the bins to p percent of the population. (a) Find the
cost to distribute bins to 25%, 50%, and 75% of the
population, then comment on the results; (b) graph
the function using an appropriate scale; and (c) use
mathematical notation to state what happens if the
city attempts to give recycling bins to 100% of the
population.
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Memory retention: Due to their asymptotic behavior,
rational functions are often used to model the mind’s ability
to retain information over a long period of time — the “use
it or lose it” phenomenon.
69. Language retention: A large group of students is
asked to memorize a list of 50 Italian words, a
language that is unfamiliar to them. The group is
then tested regularly to see how many of the words
are retained over a period of time. The average
number of words retained is modeled by the
6t 40
function W1t2 , where W(t) represents
t
the number of words remembered after t days.
a. Graph the function over the interval t 僆 3 1, 404 .
How many days until only half the words are
remembered? How many days until only onefifth of the words are remembered?
b. After 10 days, what is the average number of
words retained? How many days until only
8 words can be recalled?
c. What is the significance of the horizontal
asymptote (what does it mean in this
context)?
70. Language retention: A similar study asked
students to memorize 50 Hawaiian words, a
language that is both unfamiliar and phonetically
foreign to them (see Exercise 69). The average
number of words retained is modeled by the
4t 20
, where W(t) represents
function W1t2 t
the number of words after t days.
a. Graph the function over the interval t 僆 31, 40 4.
How many days until only half the words are
remembered? How does this compare to
Exercise 69? How many days until only onefifth of the words are remembered?
b. After 7 days, what is the average number of
words retained? How many days until only
5 words can be recalled?
c. What is the significance of the horizontal
asymptote (what does it mean in this
context)?
Concentration and dilution: When
antifreeze is mixed with water, it
becomes diluted — less than 100%
antifreeze. The more water added,
the less concentrated the antifreeze
becomes, with this process continuing until a desired
concentration is met. This application and many similar to
it can be modeled by rational functions.
Section 4.4 Graphing Rational Functions
369
71. Concentration of antifreeze: A 400-gal tank
currently holds 40 gal of a 25% antifreeze
solution. To raise the concentration of the
antifreeze in the tank, x gal of a 75% antifreeze
solution is pumped in.
a. Show the formula for the resulting
40 3x
concentration is C1x2 after
160 4x
simplifying, and graph the function over the
interval x 僆 30, 360 4 .
b. What is the concentration of the antifreeze in the
tank after 10 gal of the new solution are added?
After 120 gal have been added? How much
liquid is now in the tank?
c. If the concentration level is now at 65%, how
many gallons of the 75% solution have been
added? How many gallons of liquid are in the
tank now?
d. What is the maximum antifreeze concentration
that can be attained in a tank of this size? What
is the maximum concentration that can be
attained in a tank of “unlimited” size?
72. Concentration of sodium chloride: A sodium
chloride solution has a concentration of 0.2 oz
(weight) per gallon. The solution is pumped into an
800-gal tank currently holding 40 gal of pure
water, at a rate of 10 gal/min.
a. Find a function A(t) modeling the amount of
liquid in the tank after t min, and a function
S(t) for the amount of sodium chloride in the
tank after t min.
b. The concentration C(t) in ounces per gallon is
S1t2
, a rational function.
measured by the ratio
A1t2
Graph the function on the interval t 僆 30, 1004.
What is the concentration level (in ounces per
gallon) after 6 min? After 28 min? How many
gallons of liquid are in the tank at this time?
c. If the concentration level is now 0.184 oz/gal,
how long have the pumps been running?
How many gallons of liquid are in the tank
now?
d. What is the maximum concentration that can
be attained in a tank of this size? What is the
maximum concentration that can be attained in
a tank of “unlimited” size?
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Average cost of manufacturing an item: The cost “C” to
manufacture an item depends on the relatively fixed costs
“K” for remaining in business (utilities, maintenance,
transportation, etc.) and the actual cost “c” of
manufacturing the item (labor and materials). For x items
the cost is C1x2 K cx. The average cost
C1x2
“A” of manufacturing an item is then A1x2 .
x
73. Manufacturing water heaters: A company that
manufactures water heaters finds their fixed costs
are normally $50,000 per month, while the cost to
manufacture each heater is $125. Due to factory
size and the current equipment, the company can
produce a maximum of 5000 water heaters per
month during a good month.
a. Use the average cost function to find the
average cost if 500 water heaters are
manufactured each month. What is the average
cost if 1000 heaters are made?
b. What level of production will bring the average
cost down to $150 per water heater?
c. If the average cost is currently $137.50, how
many water heaters are being produced that
month?
d. What’s the significance of the horizontal
asymptote for the average cost function (what
does it mean in this context)? Will the
company ever break the $130 average cost
level? Why or why not?
74. Producing biodegradable disposable diapers:
An enterprising company has finally developed a
better disposable diaper that is biodegradable. The
brand becomes wildly popular and production is
soaring. The fixed cost of production is $20,000
per month, while the cost of manufacturing is
$6.00 per case (48 diapers). Even while working
three shifts around-the-clock, the maximum
production level is 16,000 cases per month. The
company figures it will be profitable if it can bring
costs down to an average of $7 per case.
a. Use the average cost function to find the
average cost if 2000 cases are produced each
month. What is the average cost if 4000 cases
are made?
b. What level of production will bring the average
cost down to $8 per case?
c. If the average cost is currently $10 per case,
how many cases are being produced?
d. What’s the significance of the horizontal
asymptote for the average cost function (what
does it mean in this context)? Will the
company ever reach its goal of $7/case at its
maximum production? What level of
production would help them meet their goal?
4–64
Test averages and grade point
averages: To calculate a test
average we sum all test points P
and divide by the number of
P
tests N: . To compute the
N
score or scores needed on future tests to raise the average
grade to a desired grade G, we add the number of additional
tests n to the denominator, and the number of additional
tests times the projected grade g on each test to the
numerator:
P ng
G1n2 . The result is a rational function with
Nn
some “eye-opening” results.
75. Computing an average grade: After four tests,
Bobby Lou’s test average was an 84.
[Hint: P 41842 336.]
a. Assume that she gets a 95 on all remaining
tests 1g 952. Graph the resulting function on
a calculator using the window n 僆 30, 20 4 and
G1n2 僆 380 to 100 4 . Use the calculator to
determine how many tests are required to lift
her grade to a 90 under these conditions.
b. At some colleges, the range for an “A” grade is
93–100. How many tests would Bobby Lou
have to score a 95 on, to raise her average to
higher than 93? Were you surprised?
c. Describe the significance of the horizontal
asymptote of the average grade function. Is a
test average of 95 possible for her under these
conditions?
d. Assume now that Bobby Lou scores 100 on all
remaining tests 1g 1002. Approximately how
many more tests are required to lift her grade
average to higher than 93?
76. Computing a GPA: At most colleges, A S 4
grade points, B S 3, C S 2, and D S 1. After
taking 56 credit hours, Aurelio’s GPA is 2.5.
[Hint: In the formula given, P 2.51562 140.]
a. Assume Aurelio is determined to get A’s
(4 grade points or g 42, for all remaining
credit hours. Graph the resulting function on a
calculator using the window n 僆 3 0, 604 and
G1n2 僆 3 2, 4 4. Use the calculator to determine
the number of credit hours required to lift his
GPA to over 2.75 under these conditions.
b. At some colleges, scholarship money is
available only to students with a 3.0 average or
higher. How many (perfect 4.0) credit hours
would Aurelio have to earn, to raise his GPA to
3.0 or higher? Were you surprised?
c. Describe the significance of the horizontal
asymptote of the GPA function. Is a GPA of
4.0 possible for him under these conditions?
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䊳
Section 4.5 Additional Insights into Rational Functions
EXTENDING THE CONCEPT
77. In addition to determining if a function has a
vertical asymptote, we are often interested in how
fast the graph approaches the asymptote. As in
previous investigations, this involves the function’s
rate of change over a small interval. Exercise 64
describes the rising cost of removing pollutants
from the air. As noted there, the rate of increase in
the cost changes as higher requirements are set. To
quantify this change, we’ll compute the rate of
C1x2 2 C1x1 2
¢C
250x
.
change
for C1x2 x2 x1
¢x
100 x
a. Find the rate of change of the function in the
following intervals:
x 僆 3 60, 614 x 僆 3 70, 71 4
x 僆 3 80, 814 x 僆 390, 91 4
b. What do you notice? How much did the rate
increase from the first interval to the second?
From the second to the third? From the third to
the fourth?
䊳
371
c. Recompute parts (a) and (b) using the function
350x
. Comment on what you notice.
C1x2 100 x
ax2 k
, where a, b,
bx2 h
k, and h are constants and a, b 7 0.
a. What can you say about asymptotes and
intercepts of this function if h, k 7 0?
b. Now assume k 6 0 and h 7 0. How does this
affect the asymptotes? The intercepts?
c. If b 1 and a 7 1, how does this affect the
results from part (b)?
d. How is the graph affected if k 7 0 and h 6 0?
e. Find values of a, b, h, and k that create a
function with a horizontal asymptote at y 32 ,
x-intercepts at 12, 02 and (2, 0), a y-intercept
of 10, 42 , and no vertical asymptotes.
78. Consider the function f 1x2 MAINTAINING YOUR SKILLS
79. (1.4) Find the equation of a line that is
perpendicular to 3x 4y 12 and contains the
point 12, 32.
80. (4.1) Use synthetic division and the remainder
theorem to find the value of f (4), f 1 32 2 , and
f (2):f 1x2 2x3 7x2 5x 3.
81. (3.2) Solve the following equation using the
quadratic formula, then write the equation in
factored form: 12x2 55x 48 0.
82. (Appendix A.1/3.1) Describe/Define each set of
numbers: complex ⺓, rational ⺡, and integers ⺪.
4.5
Additional Insights into Rational Functions
LEARNING OBJECTIVES
In Section 4.5 you will see
how we can:
A. Graph rational functions with
removable discontinuities
B. Graph rational functions
with oblique or nonlinear
asymptotes
C. Solve applications involving
rational functions
In Section 4.4, we studied rational functions whose graphs had horizontal and/or vertical asymptotes. In this section, we’ll study functions with asymptotes that are neither
horizontal nor vertical. In addition, we’ll further explore the “break” we saw in graphs
of certain piecewise-defined functions, that of a simple “hole” created when the numerator and denominator of a rational function share a common variable factor.
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CHAPTER 4 Polynomial and Rational Functions
A. Rational Functions and Removable Discontinuities
In Example 6 of Section 2.5, we graphed the piecewise-defined function
x2 4
x2
h1x2 • x 2
. The second piece is simply the point (2, 1). The first piece
1
x2
is a rational function, but instead of a vertical asymptote at x 2 (the zero of the
denominator), its graph is actually the line y x 2 with a “hole” at (2, 4) called a
removable discontinuity (Figure 4.60). The hole occurs because the numerator and
x2 4
denominator of y share the common factor 1x 22 , and canceling these
x2
factors leaves y x 2, a continuous function. However, the original function is not
defined at x 2, so we must delete the single point at x 2, y 4 from the graph of
the line (Figure 4.60).
Figure 4.60
Figure 4.61
y
y
5
5
4
4
(2, 4)
3
2
h(x)
2
H(x)
(2, 1)
1
5 4 3 2 1
1
(2, 4)
3
1
2
3
4
5
x
1
5 4 3 2 1
1
2
2
3
3
4
4
5
5
1
2
3
4
5
x
We can remove or fix this discontinuity by redefining the second piece
as h1x2 4, when x 2. This would create a new and continuous function,
x2 4
x2
H1x2 • x 2
(Figure 4.61).
4
x2
It’s possible for a rational graph to have more than one removable discontinuity, or
to be nonlinear with a removable discontinuity. For cases where we elect to repair the
break, we will adopt the convention of using the corresponding upper case letter to
name the new function, as we did here.
EXAMPLE 1
䊳
Graphing Rational Functions with Remo vable Discontinuities
x3 8
. If there is a removable discontinuity, repair the
x2
break using an appropriate piecewise-defined function.
Graph the function t1x2 Solution
䊳
Note the domain of t does not include x 2. We begin by factoring as before to
identify zeroes and asymptotes, but find the numerator and denominator share a
common factor, which we remove.
t1x2 x3 8
x2
1x 22 1x2 2x 42
x2
x 2x 4; where x 2
2
The graph of t will be the same as y x2 2x 4 for all values except x 2.
Here we have a parabola, opening upward, with y-intercept (0, 4). From the vertex
122
b
1, giving y 3
formula, the x-coordinate of the vertex will be
2a
2112
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Section 4.5 Additional Insights into Rational Functions
after substitution. The vertex is (1, 3). Evaluating
t1⫺12 we find 1⫺1, 72 is on the graph, giving the
point (3, 7) using the axis of symmetry. We draw
a parabola through these points, noting the
original function is not defined at ⫺2, and there
will be a “hole” in the graph at 1⫺2, y2. The
value of y is found by substituting ⫺2 for x in the
simplified form: 1⫺22 2 ⫺ 21⫺22 ⫹ 4 ⫽ 12. This
information produces the graph shown. We can
repair the break using the function
x3 ⫹ 8
x ⫽ ⫺2
T1x2 ⫽ • x ⫹ 2
12
x ⫽ ⫺2
t(x)
(4, 12)
from
symmetry
(⫺2, 12)
10
(3, 7)
from
symmetry
(⫺1, 7)
5
(1, 3)
x
5
axis of
symmetry
Now try Exercises 7 through 18
The
䊳
feature of a graphing calculator is a wonderful tool for understanding the
f 1x2
, where
characteristics of a function. We’ll illustrate using the function h1x2 ⫽
g1x2
TRACE
f 1x2 ⫽ x3 ⫺ 2x2 ⫺ 3x ⫹ 6 and g1x2 ⫽ x ⫺ 2 (similar to Example 1). Enter
A. You’ve just seen how
we can graph rational
functions with removable
discontinuities
the Y= screen as Y1 (Figure 4.62), then graph
the function using ZOOM 4:ZDecimal. Recall
this will allow the calculator to trace through
¢x intervals of 0.1. After pressing the TRACE
key, the cursor appears on the graph at the
y-intercept 10, ⫺32 and its location is displayed
at the bottom of the screen. Note that there is a
“hole” in the graph in the first quadrant (Figure 4.63). We can walk the cursor along the
curve in either direction using the left arrow
and right arrow
keys to determine exactly
where this hole occurs. Walking the cursor to
the right, we note that no output is displayed
for x ⫽ 2.
f 1x2
Next, simplify
and enter the result
g1x2
as Y2. Factoring the numerator by grouping
and reducing the common factors gives
1x2 ⫺ 32 1x ⫺ 22
f 1x2
⫽
, so Y2 ⫽ x2 ⫺ 3.
g1x2
1x ⫺ 22
Graphing both functions reveals that they are
identical, except that Y2 ⫽ x2 ⫺ 3 covers the
hole left by Y1 using x ⫽ 2, y ⫽ 1. In other
words, Y1 is equivalent to Y2 except at x ⫽ 2.
This can also be seen using the TABLE feature
of a calculator, which displays an error message
for Y1 when x ⫽ 2 is input, but shows an output
of 1 for Y2 (Figure 4.64). The bottom line is—
the domain of h is all real numbers except x ⫽ 2.
f 1x2
g1x2
on
Figure 4.62
Figure 4.63
3.1
⫺4.7
4.7
⫺3.1
Figure 4.64
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CHAPTER 4 Polynomial and Rational Functions
B. Rational Functions with Oblique or Nonlinear Asymptotes
In Section 4.4, we found that for V1x2 p1x2
, the location of nonvertical asymptotes
d1x2
was determined by comparing the degree of p with the degree of d. As review, for p(x)
with leading term axn and d(x) with leading term bxm,
• If n m, the line y ab
is a horizontal asymptote.
• If n 6 m, the line y 0
is a horizontal asymptote.
But what happens if the degree of the numerator is greater than the degree of the denominator? To investigate, consider the functions f, g, and h in Figures 4.65 to 4.68,
whose only difference is the degree of the numerator.
Figure 4.65
f 1x2 Figure 4.66
g 1x2 2x
x 1
y
2
Figure 4.67
2x2
h 1x2 x 1
y
2
5
5
5
4
4
4
3
3
3
2
2
2
1
1
1
5 4 3 2 1
1
1
2
3
4
5
x
5 4 3 2 1
1
1
2
3
4
5
x
5 4 3 2 1
1
2
2
2
3
3
3
4
4
4
5
5
5
2x3
x 1
y
2
1
2
3
4
5
x
The graph of f has a horizontal asymptote at y 0 since the denominator is of larger
degree (as 冟x冟 S q, y S 02. As we might have anticipated, the horizontal asymptote
for g is y 2, the ratio of leading coefficients
Figure 4.68
(as 冟x冟 S q, y S 22. The graph of h has no horY2 2X
izontal asymptote, yet appears to be asymptotic
to some slanted line. The table in Figure 4.68
suggests that as 冟x冟 S q, y S 2x Y2. To see
2x3
why, note the function h1x2 2
can be
x 1
considered an “improper fraction,” similar to
how we apply this designation to the fraction 32.
To write h in “proper” form, we use long
division, writing the dividend as 2x3 0x2 0x 0, and the divisor as x2 0x 1.
2x3 from dividend
The ratio 2
shows 2x will be our first multiplier.
x from divisor
2x
divisor
S x2 0x 1冄 2x3 0x2 0x 0
12x3 0x2 2x2
2x
multiply 2x 1x 2 0x 12
subtract, next term is 0
2x
2x
. Note as 冟x冟 S q, the term 2
becomes
x 1
x 1
very small and closer to zero, so h1x2 ⬇ 2x for large x. This is an example of an
oblique asymptote. In general,
The result shows h1x2 2x 2
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375
Section 4.5 Additional Insights into Rational Functions
Oblique and Nonlinear Asymptotes
Given V1x2 p1x2
is a rational function in simplest form, where the degree of p is
d1x2
greater than the degree of d, the graph will have an oblique or nonlinear asymptote as
determined by q(x), where q(x) is the quotient polynomial after division.
If the denominator is a monomial, term-by-term division is the most efficient
means of computing the quotient. If the denominator is not a monomial, either synthetic division or long division must be used. From our work in Section 4.4, q(x) will
give the location of the asymptote, and the zeroes of the remainder (if they exist) will
indicate where the function crosses the asymptote.
We conclude that an oblique or slant asymptote occurs when the degree of the numerator is one more than the degree of the denominator, as that indicates q(x) will be linear. A
nonlinear asymptote will occur when the degree of the numerator is larger by two or more.
EXAMPLE 2
䊳
Graphing a Rational Function with an Oblique Asymptote
Graph the function f 1x2 Solution
䊳
x2 1
.
x
1x 12 1x 12
and proceed:
x
y-intercept: The graph has no y-intercept since f (0) is undefined.
x-intercepts: From 1x 12 1x 12 0, the x-intercepts are 11, 02 and (1, 0).
Since both have multiplicity 1, the graph will cross the x-axis and the function
will change sign at these points.
Vertical asymptote(s): x 0 with multiplicity 1. The function will change sign
at x 0.
Horizontal/oblique asymptote: Since the degree of numerator 7 the degree of
denominator, we rewrite f using division. Using term-by-term division (the
x2 1
x2
1
1
denominator is a monomial) produces f 1x2 x . The
x
x
x
x
quotient polynomial is q1x2 x and the graph has the oblique asymptote y x.
To determine if the function will cross the asymptote, we note the remainder is
1
r1x2 , which has no real zeroes. The graph will not cross the oblique
x
asymptote.
Using the Guidelines, we find f 1x2 1.
2.
3.
4.
5.
The information from steps 1 through 5 is displayed in Figure 4.69. While this is
sufficient to sketch the graph, we select x 4 and x 4 and compute the
15
additional points: f 142 15
4 and f 142 4 . To meet all necessary conditions,
we complete the graph as shown in Figure 4.70.
Figure 4.69
y
10
x0
Figure 4.70
y
yx
10
冢4, 154冣
pos
10
neg
neg pos
10
yx
10
x
10
冢4,
10
x
冣
15
4
10
Now try Exercises 19 through 24
䊳
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CHAPTER 4 Polynomial and Rational Functions
EXAMPLE 3
䊳
Graphing a Rational Function with an Oblique Asymptote
x2
x1
Graph the function: h1x2 Solution
䊳
The function is already in “factored form.”
1. y-intercept: Since h102 0, the y-intercept is (0, 0).
2. x-intercept: (0, 0); From, x2 0, we have x 0 with multiplicity two. The
x-intercept is (0, 0) and the function will not change sign here.
3. Vertical asymptote: Solving x 1 0 gives x 1 with multiplicity one.
There is a vertical asymptote at x 1 and the function will change sign here.
4. Horizontal/oblique asymptote: Since the degree of numerator 7 the degree of
denominator, we rewrite h using division. The denominator is linear so we use
synthetic division:
use 1 as a “divisor”
1
1
T
1
0
1
1
0
1
1
coefficients of dividend
quotient and remainder
Since q1x2 x 1 the graph has an oblique asymptote at y x 1.
5. The remainder is r 1x2 1, a constant. The graph will not cross the slant
asymptote.
The information gathered in steps 1 through 5 is shown Figure 4.71, and is actually
sufficient to complete the graph. If you feel a little unsure about how to “puzzle”
out the graph, find additional points in the first and third quadrants: h122 4 and
h122 43. Since the graph will “bounce” at x 0 and output values must
change sign at x 1, all conditions are met with the graph shown in Figure 4.72.
Figure 4.71
Figure 4.72
y
x1
10
y
10
yx1
yx1
(2, 4)
冢2, 43冣
neg
n
e
g
10
pos
10
x
x1
10
10
10
x
10
Now try Exercises 25 through 46
B. You’ve just seen how
we can graph rational
functions with oblique or
nonlinear asymptotes
Finally, it would be a mistake to think that all
asymptotes are linear. In fact, when the degree of the
numerator is two more than the degree of the denominator, a parabolic asymptote results. Functions of this
type often occur in applications of rational functions,
and are used to minimize cost, materials, distances, or
other considerations of great importance to business and
x4 1
, term-by-term division
industry. For f 1x2 x2
1
gives x2 2 and the quotient q 1x2 x2 is a nonlinear,
x
parabolic asymptote (see Figure 4.73). For more on
nonlinear asymptotes, see Exercises 47 through 50.
䊳
Figure 4.73
y
y x2
10
f(x)
(1, 2)
(1, 2)
5
5
10
x0
x
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Section 4.5 Additional Insights into Rational Functions
377
C. Applications of Rational Functions
Rational functions have applications in a wide variety of fields, including environmental studies, manufacturing, and various branches of medicine. In most practical
applications, only the values from Quadrant I have meaning since inputs and outputs
must often be positive (see Exercises 51 and 52). Here we investigate an application
involving manufacturing and average cost.
EXAMPLE 4
䊳
Solving an Application of Rational Functions
Suppose the cost (in thousands of dollars) of manufacturing x thousand of a given
item is modeled by the function C1x2 x2 4x 3. The average cost of each
item would then be expressed by
A1x2 total cost
x2 4x 3
x
number of items
a. Graph the function A(x).
b. Find how many thousand items are manufactured when the average cost is $8.
c. Determine how many thousand items should be manufactured to minimize the
average cost (use the graph to estimate this minimum average cost).
Solution
䊳
a. The function is already in simplest form.
1. y-intercept: none 3 A102 is undefined]
2. x-intercept(s): After factoring we obtain 1x 32 1x 12 0, and the zeroes of
the numerator are x 1 and x 3, both with multiplicity one. The graph
will cross the x-axis at each intercept.
3. Vertical asymptote: x 0, multiplicity one; the function will change sign
at x 0.
4. Horizontal/oblique asymptote: The degree of numerator 7 the degree of
denominator, so we divide using term-by-term division:
x2 4x 3
x2
4x
3
x
x
x
x
3
x4
x
The line q1x2 x 4 is an oblique asymptote.
3
5. The remainder has no real zeroes, so the graph will not cross the slant
x
asymptote.
The function changes sign at both x-intercepts and at the asymptote x 0. The
information from steps 1 through 5 is shown in Figure 4.74 and perhaps an
additional point in Quadrant I would help to complete the graph: A112 8. The
point (1, 8) is on the graph, showing A is positive in the interval containing 1 (since
y 8 is positive). Since output values will alternate in sign as stipulated above, all
conditions are met with the graph shown in Figure 4.75.
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CHAPTER 4 Polynomial and Rational Functions
Figure 4.74
Figure 4.75
y
y
(1, 8)
(1, 8)
y ⫽x ⫹ 4
y ⫽x ⫹ 4
5
5
pos
⫺5
neg
pos
n
e
g
⫺5
5
⫺5
x
5
⫺5
x⫽0
x
x⫽0
b. To find the number of items manufactured when average cost is $8, we replace
x2 ⫹ 4x ⫹ 3
A(x) with 8 and solve:
⫽ 8:
x
x2 ⫹ 4x ⫹ 3 ⫽ 8x
x2 ⫺ 4x ⫹ 3 ⫽ 0
1x ⫺ 12 1x ⫺ 32 ⫽ 0
x ⫽ 1 or x ⫽ 3
The average cost is $8 when 1000 items or 3000 items are manufactured.
c. From the graph, it appears that the minimum average cost is close to $7.50,
when approximately 1500 to 1800 items are manufactured. Using a graphing
calculator, we find that the minimum average cost is approximately $7.46,
when about 1732 items are manufactured (Figure 4.76).
Figure 4.76
12
⫺10
10
⫺8
Now try Exercises 55 and 56
䊳
In some applications, the functions we use are initially defined in two variables
rather than just one, as in H1x, y2 ⫽ 1x ⫺ 502 1y ⫺ 802. However, in the solution
process a substitution is used to rewrite the relationship as a function in one variable
and we can proceed as before.
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379
Section 4.5 Additional Insights into Rational Functions
EXAMPLE 5
䊳
Using a Rational Function to Solve a La yout Application
rear fence line
The building codes in a new subdivision
require that a rectangular home be built at
40 ft
least 20 ft from the street, 40 ft from the
neighboring lots, and 30 ft from the rear
fence line.
a. Find a function A(x, y) for the area of
the lot, and a function H(x, y) for the
20 ft
30 ft
area of the home (the inner rectangle). y
b. If a new home is to have a floor area
of 2000 ft2, H1x, y2 2000. Substitute
2000 for H(x, y) and solve for y, then
substitute the result in A(x, y) to write
the area A as a function of x alone
40 ft
(simplify the result).
c. Graph A(x) on a calculator, using the
x
window X 僆 350, 150 4;
Y 僆 330,000, 30,000 4 . Then graph y 80x 2000 on the same screen.
How are these two graphs related?
d. Use the graph of A(x) in Quadrant I to determine the minimum dimensions of a
lot that satisfies the subdivision’s requirements (to the nearest tenth of a foot).
Also state the dimensions of the house.
Solution
䊳
a. The area of the lot is simply width times length, so A1x, y2 xy. For the
house, these dimensions are decreased by 50 ft and 80 ft respectively, so
H1x, y2 1x 502 1y 802.
b. Given H1x, y2 2000 produces the equation 2000 1x 5021y 802, and
solving for y gives
2000 1x 502 1y 802
2000
y 80
x 50
2000
80 y
x 50
801x 502
2000
y
x 50
x 50
80x 2000
y
x 50
given equation
divide by x 50
add 80
find LCD
combine terms
Substituting this expression for y in A1x, y2 xy produces
80x 2000
b
x 50
80x2 2000x
x 50
A1x2 xa
c. The graph of Y1 A1x2 appears in
Figure 4.77 using the prescribed
window. Y2 80x 2000 appears
to be an oblique asymptote for A,
which can be verified using synthetic
division.
substitute
80x 2000
for y
x 50
multiply
Figure 4.77
30,000
50
150
30,000
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CHAPTER 4 Polynomial and Rational Functions
d. Using the 2nd TRACE (CALC)
3:minimum feature of a calculator,
the minimum width is x ⬇ 85.4 ft
(see Figure 4.78). Substituting 85.4 for
80x 2000
, gives the
x in y x 50
length y ⬇ 136.5 ft. The dimensions
of the house must be
85.4 50 35.4 ft, by
136.5 80 56.5 ft.
Figure 4.78
30,000
50
150
30,000
As expected, the area of the house will be 135.42 156.52 ⬇ 2000 ft2.
C. You’ve just seen how
we can solve applications
involving rational functions
Now try Exercises 57 through 60
䊳
4.5 EXERCISES
䊳
CONCEPTS AND VOCABULAR Y
Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.
3x3
will have a _________
x2 4
asymptote, since the degree of the numerator is one
greater than the degree of the denominator.
䊳
1. The graph of V1x2 2. If the degree of the numerator is greater than the
degree of the denominator, the graph will have an
_________ or _________ asymptote.
3. If the degree of the numerator is _________ more
than the degree of the denominator, the graph will
have a parabolic asymptote.
4. If the denominator is a _________, use term by
term division to find the quotient. Otherwise,
_________ or long division must be used.
5. Discuss/Explain how you would create a function
with a parabolic asymptote and two vertical
asymptotes.
6. Complete Exercise 7 in expository form. That is,
work this exercise out completely, discussing each
step of the process as you go.
DEVELOPING YOUR SKILLS
Graph each function. If there is a removable
discontinuity, repair the break using an appropriate
piecewise-defined function.
x2 4
7. f 1x2 x2
9. g1x2 11. h1x2 x2 9
8. f 1x2 x3
x2 2x 3
x1
10. g1x2 x2 3x 10
x5
3x 2x2
2x 3
12. h1x2 4x 5x2
5x 4
17. r 1x2 x3 3x2 x 3
x2 2x 3
18. r 1x2 x3 2x2 4x 8
x2 4
Graph each function using the Guidelines for Graphing
Rational Functions, which is simply modified to include
nonlinear asymptotes. Clearly label all intercepts and
asymptotes and any additional points used to sketch the
graph. Round to tenths as needed.
x2 4
x
x3 8
13. p1x2 x2
8x3 1
14. p1x2 2x 1
19. Y1 x3 7x 6
15. q1x2 x1
x3 3x 2
16. q1x2 x2
21. v1x2 3 x2
x
20. Y2 x2 x 6
x
22. V1x2 7 x2
x
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Section 4.5 Additional Insights into Rational Functions
23. w1x2 x2 1
x
25. h1x2 x3 2x2 3
x3 x2 2
26. H1x2 2
x
x2
27. Y1 x3 3x2 4
x2
29. f 1x2 31. Y3 x3 3x 2
x2
x3 5x2 4
x2
28. Y2 32. Y4 x3 x2 4x 4
x2
34. R1x2 x3 2x2 9x 18
x2
35. g1x2 x2 4x 4
x3
39. Y3 䊳
x2 4
x1
x3 3x2 4
x2
30. F1x2 33. r1x2 x2 1
37. f 1x2 x1
x2 4
2x
24. W1x2 x3 12x 16
x2
x3 5x2 6
x2
41. v1x2 x3 4x
x2 1
42. V1x2 9x x3
x2 4
43. w1x2 16x x3
x2 4
44. W1x2 x3 7x 6
2 x2
45. Y1 x3 3x 2
x2 9
46. Y2 x3 x2 12x
x2 7
47. p1x2 x4 4
x2 1
49. q1x2 10 9x2 x4
x4 2x2 3
50.
Q1x2
x2 5
x2
48. P1x2 x4 5x2 4
x2 2
Graph each function and its nonlinear asymptote on the
same screen, using the window specified. Then locate the
minimum value of f in the first quadrant.
36. G1x2 x2 2x 1
x2
x2 x 1
38. F1x2 x1
40. Y4 381
x2 x 6
x1
x3 500
;
x
X 僆 324, 24 4, Y 僆 3500, 5004
51. f 1x2 2␲x3 750
;
x
X 僆 312, 124 , Y 僆 3750, 7504
52. f 1x2 WORKING WITH FORMULAS
53. Area of a first quadrant triangle:
1 ka2
b
A1a2 ⴝ a
2 aⴚh
y
The area of a right triangle in
(0, y)
the first quadrant, formed by a
line with negative slope
(h, k)
through the point (h, k) and
legs that lie along the positive
axes is given by the formula
(a, 0)
shown, where a represents the
x
x-intercept of the resulting
line 1h 6 a2. The area of the triangle varies with
the slope of the line. Assume the line contains the
point (5, 6).
a. Find the equation of the vertical and slant
asymptotes.
b. Find the area of the triangle if it has an
x-intercept of (11, 0).
c. Use a graphing calculator to graph the function
on an appropriate window. Does the shape of
the graph look familiar? Use the calculator to
find the value of a that minimizes A(a). That is,
find the x-intercept that results in a triangle
with the smallest possible area.
54. Surface area of a cylinder with fixed volume:
2␲r 3 ⴙ 2V
Sⴝ
r
It’s possible to construct
750 cm3
many different cylinders that
750 cm3
will hold a specified volume,
by changing the radius and
height. This is critically important to producers
who want to minimize the cost of packing canned
goods and marketers who want to present an
attractive product. The surface area of the cylinder
can be found using the formula shown, where the
radius is r and V ␲r2h is known. Assume the
fixed volume is 750 cm3.
a. Find the equation of the vertical asymptote. How
would you describe the nonlinear asymptote?
b. If the radius of the cylinder is 2 cm, what is its
surface area?
c. Use a graphing calculator to graph the function
on an appropriate window, and use it to find
the value of r that minimizes S(r). That is, find
the radius that results in a cylinder with the
smallest possible area, while still holding a
volume of 750 cm3.
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䊳
CHAPTER 4 Polynomial and Rational Functions
4–76
APPLICATIONS
Costs of manufacturing: As in Example 4, the cost C(x) of
manufacturing is sometimes nonlinear and can increase
dramatically with each item. For the average
C1x2
cost function A1x2 , consider the following.
x
55. Assume the monthly cost of manufacturing
custom-crafted storage sheds is modeled by the
function C1x2 4x2 53x 250.
a. Write the average cost function and state the
equation of the vertical and oblique
asymptotes.
b. Enter the cost function C(x) as Y1 on a
graphing calculator, and the average cost
function A(x) as Y2. Using the TABLE feature,
find the cost and average cost of making 1, 2,
and 3 sheds.
c. Scroll down the table to where it appears that
average cost is a minimum. According to the
table, how many sheds should be made each
month to minimize costs? What is the
minimum cost?
d. Graph the average cost function and its
asymptotes, using a window that shows the
entire function. Use the graph to confirm the
result from part (c).
56. Assume the monthly cost of manufacturing
playground equipment that combines a play house,
slides, and swings is modeled by the function
C1x2 5x2 94x 576. The company has
projected that they will be profitable if they can
bring their average cost down to $200 per set of
playground equipment.
a. Write the average cost function and state the
equation of the vertical and oblique asymptotes.
b. Enter the cost function C(x) as Y1 on a
graphing calculator, and the average cost
function A(x) as Y2. Using the TABLE feature,
find the cost and average cost of making 1, 2,
and 3 playground equipment combinations.
Why would the average cost fall so
dramatically early on?
c. Scroll down the table to where it appears that
average cost is a minimum. According to the
table, how many sets of equipment should be
made each month to minimize costs? What is
the minimum cost? Will the company be
profitable under these conditions?
d. Graph the average cost function and its
asymptotes, using a window that shows the
entire function. Use the graph to confirm the
result from part (c).
Minimum cost of packaging: Similar to Exercise 54,
manufacturers can minimize their costs by shipping
merchandise in packages that use a minimum amount of
material. After all, rectangular boxes come in different sizes
and there are many combinations of length, width, and
height that will hold a specified volume.
57. A clothing manufacturer wishes
to ship lots of 12 ft3 of clothing in
boxes with square ends and
x
y
rectangular sides.
x
a. Find a function S(x, y) for the
surface area of the box, and a function V(x, y)
for the volume of the box.
b. Solve for y in V1x, y2 12 (volume is 12 ft3 2
and use the result to write the surface area as a
function S(x) in terms of x alone (simplify the
result).
c. On a graphing calculator, graph the function
S(x) using the window x 僆 38, 8 4;
y 僆 3100, 100 4. Then graph y 2x2 on the
same screen. How are these two graphs
related?
d. Use the graph of S(x) in Quadrant I to
determine the dimensions that will minimize
the surface area of the box, yet still hold 12 ft3
of clothing. Clearly state the values of x and y,
in terms of feet and inches, rounded to the
nearest 21 in.
58. A maker of packaging materials needs to ship
36 ft3 of foam “peanuts” to his customers across
the country, using boxes with the
dimensions shown.
a. Find a function S1x, y2 for
x
the surface area of the box,
y
and a function V1x, y2 for the
x2
volume of the box.
b. Solve for y in V1x, y2 36 (volume is 36 ft3 2 ,
and use the result to write the surface area as a
function S(x) in terms of x alone (simplify the
result).
c. On a graphing calculator, graph the function
S(x) using the window
x 僆 310, 104 ; y 僆 3200, 2004 . Then graph
y 2x2 4x on the same screen. How are
these two graphs related?
d. Use the graph of S(x) in Quadrant I to
determine the dimensions that will minimize
the surface area of the box, yet still hold the
foam peanuts. Clearly state the values of x and
y, in terms of feet and inches, rounded to the
nearest 12 in.
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383
Section 4.5 Additional Insights into Rational Functions
Printing and publishing: In the design of magazine pages, posters, and other published materials, an effort is made to
maximize the usable area of the page while maintaining an attractive border, or minimizing the page size that will hold a
certain amount of print or art work.
59. An editor has a story that requires 60 in2 of print. Company standards
require a 1-in. border at the top and bottom of a page, and 1.25-in. borders
along both sides.
a. Find a function A(x, y) for the area of the page, and a function R(x, y)
for the area of the inner rectangle (the printed portion).
b. Solve for y in R1x, y2 60, and use the result to write the area from
part (a) as a function A(x) in terms of x alone (simplify the result).
y
c. On a graphing calculator, graph the function A(x) using the window
x 僆 3 30, 30 4 ; y 僆 3100, 200 4. Then graph y 2x 60 on the same
screen. How are these two graphs related?
d. Use the graph of A(x) in Quadrant I to determine the page of minimum
size that satisfies these border requirements and holds the necessary
print. Clearly state the values of x and y, rounded to the nearest
hundredth of an inch.
60. The Poster Shoppe creates posters, handbills, billboards, and other
advertising for business customers. An order comes in for a poster with
500 in2 of usable area, with margins of 2 in. across the top, 3 in. across the
bottom, and 2.5 in. on each side.
a. Find a function A(x, y) for the area of the page, and a function R(x, y)
for the area of the inner rectangle (the usable area).
b. Solve for y in R1x, y2 500, and use the result to write the area from
part (a) as a function A(x) in terms of x alone (simplify the result).
c. On a graphing calculator, graph A(x) using the window
x 僆 3 100, 1004 ; y 僆 3800, 1600 4. Then graph y 5x 500 on the
same screen. How are these two graphs related?
d. Use the graph of A(x) in Quadrant I to determine the poster of
minimum size that satisfies these border requirements and has the
necessary usable area. Clearly state the values of x and y, rounded to
the nearest hundredth of an inch.
1 in.
1~ in.
1~ in.
1 in.
x
2 in.
2q in.
2q in.
y
3 in.
x
61. The formula from Exercise 54 has an interesting derivation. The volume of a cylinder is V ␲r2h, while the
surface area is given by S 2␲r2 2␲rh (the circular top and bottom the area of the side).
a. Solve the volume formula for the variable h.
b. Substitute the resulting expression for h into the surface area formula and simplify.
c. Combine the resulting two terms using the least common denominator, and the result is the formula from
Exercise 54.
d. Assume the volume of a can must be 1200 cm3. Use a calculator to graph the function S using an
appropriate window, then use it to find the radius r and height h that will result in a cylinder with the
smallest possible area, while still holding a volume of 1200 cm3. What is the minimum surface area?
Also see Exercise 62.
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CHAPTER 4 Polynomial and Rational Functions
62. The surface area of a spherical
h
cap is given by S 2␲rh,
where r is the radius of the
r
sphere and h is the
perpendicular distance from
the sphere’s surface to the
plane intersecting the sphere,
forming the cap. The volume of the cap is
V 13␲h2 13r h2. Similar to Exercise 61, a
formula can be found that will minimize the area of
a cap that holds a specified volume.
a. Solve the volume formula for the variable r.
䊳
EXTENDING THE CONCEPT
63. Consider rational functions of the form
x2 a
f 1x2 . Use a graphing calculator to
xb
explore cases where a b2 1, a b2, and
a b2 1. What do you notice? Explain/Discuss
why the graphs differ. It’s helpful to note that when
graphing functions of this form, the “center” of the
graph will be at 1b, b2 a2, and the window size
can be set accordingly for an optimal view. Do
some investigation on this function and
determine/explain why the “center” of the graph is
at 1b, b2 a2.
64. The formula from Exercise 53 also has an
interesting derivation, and the process involves
this sequence:
y
(0, y)
a. Use the points (a, 0) and
(h, k) to find the slope of
(h, k)
the line, and the pointslope formula to find the
equation of the line in
(a, 0)
terms of y.
䊳
b. Substitute the resulting expression for r into
the surface area formula and simplify. The
result is a formula for surface area given solely
in terms of the volume V and the height h.
c. Assume the volume of the spherical cap is
500 cm3. Use a graphing calculator to graph
the resulting function on an appropriate
window, and use the graph to find the height h
that will result in a spherical cap with the
smallest possible area, while still holding a
volume of 500 cm3.
d. Use this value of h and V 500 cm3 to find
the radius of the sphere.
b. Use this equation to find the x- and y-intercepts
of the line in terms of a, k, and h.
c. Complete the derivation using these intercepts
and the triangle formula A 12BH.
d. If the lines goes through (4, 4) the area formula
1 4a2
b. Find the minimum
becomes A a
2 a4
value of this rational function. What can you
say about the triangle with minimum area
through (h, k), where h k? Verify using the
points (5, 5), and (6, 6).
65. Referring to Exercises 54 and 61, suppose that
instead of a closed cylinder, with both a top and
bottom, we needed to manufacture open cylinders,
like tennis ball cans that use a lid made from a
different material. Derive the formula that will
minimize the surface area of an open cylinder, and
use it to find the cylinder with minimum surface
area that will hold 90 in3 of material.
x
MAINTAINING YOUR SKILLS
5i
, then check
1 2i
your answer using multiplication.
66. (3.1) Compute the quotient
67. (1.4) Write the equation of the line in slope
intercept form and state the slope and y-intercept:
3x 4y 16.
68. (3.2) Given f 1x2 ax bx c, use the
discriminant to state conditions where the function
will have: (a) two, real/rational roots, (b) two,
real/irrational roots, (c) one real and rational root,
(d) one real/irrational root, (e) one complex root,
and (f) two complex roots.
69. (Appendix A.6/3.2) For triangle ABC as shown, (a)
find the perimeter; (b) find the length of CD, given
1CB2 2 AB # DB; (c) find the area; and (d) find the
areas of the two smaller triangles.
C
2
12 cm
A
5 cm
D
B
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Precalculus—
4.6 Polynomial and Rational Inequalities
LEARNING OBJECTIVES
A. Polynomial Inequalities
In Section 4.6 you will see
how we can:
Our work with quadratic inequalities (Section 3.4) transfers seamlessly to inequalities
involving higher degree polynomials and the same two methods can be employed. The
first involves drawing a quick sketch of the function, and using the concepts of multiplicity and end-behavior. The second involves the use of multiple interval tests, to
check on the sign of the function in each interval.
A. Solve polynomial
inequalities
B. Solve rational inequalities
C. Solve applications of
inequalities
Solving Inequalities Graphically
After writing the polynomial in standard form, find the zeroes, plot them on the x-axis,
and determine the solution set using end-behavior and the behavior at each zero
(cross—sign change; or bounce—no change in sign). In this process, any irreducible
quadratic factors can be ignored, as they have no effect on the solution set. In summary,
Solving Polynomial Inequalities
Given f(x) is a polynomial in standard form,
1. Write f in completely factored form.
2. Plot real zeroes on the x-axis, noting their multiplicity.
• If the multiplicity is odd the function will change sign.
• If the multiplicity is even, there will be no change in sign.
3. Use the end-behavior to determine the sign of f in the outermost intervals,
then label the other intervals as f 1x2 6 0 or f 1x2 7 0 by analyzing the
multiplicity of neighboring zeroes.
4. State the solution in interval notation.
EXAMPLE 1
䊳
Solving a Polynomial Inequality
Solve the inequality x3 18 6 4x2 3x.
Solution
䊳
In standard form we have x3 4x2 3x 18 6 0, which is equivalent to
f 1x2 6 0 where f 1x2 x3 4x2 3x 18. The polynomial cannot be factored
by grouping and testing 1 and 1 shows neither is a zero. Using x 2 and
synthetic division gives
use 2 as a “divisor”
2
1
3
12
9
4
2
6
↓
1
18
18
0
with a quotient of x 6x 9 and a remainder of zero.
1. The factored form is f 1x2 1x 22 1x2 6x 92 1x 221x 32 2.
2. The graph will bounce off the x-axis at x 3 ( f will not change sign), and
cross the x-axis at x 2 ( f will change sign). This is illustrated in Figure 4.79,
which uses open dots due to the strict inequality.
2
Figure 4.79
no change
4
3
2
change
1
0
1
2
3
4 x
3. The polynomial has odd degree with a positive lead coefficient, so endbehavior is down/up, which we note in the outermost intervals. Working
from the left, f will not change sign at x 3, showing f 1x2 6 0 in the left
and middle intervals. This is supported by the y-intercept (0, 18). See
Figure 4.80.
4–79
385
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4–80
CHAPTER 4 Polynomial and Rational Functions
Figure 4.80
no change
4
f(x) 0
3
end-behavior
is “up”
change
2
1
0
2
1
f(x) 0
3
4
f(x) 0
f(0) 18
end-behavior
is “down”
4. From the diagram, we see that f 1x2 6 0
for x 僆 1q, 32 ´ 13, 22 , which
must also be the solution interval for
x3 18 6 4x2 3x. The complete
graph appearing in Figure 4.81 definitely
shows the graph is below the x-axis
3 f 1x2 6 0 4 from q to 2, except at
x 3 where the graph touches the
x-axis without crossing.
Figure 4.81
30
5
5
30
Now try Exercises 7 through 18
EXAMPLE 2
䊳
䊳
Solving a Polynomial Inequality
Solve the inequality x4 4x 9x2 12.
Solution
䊳
Writing the polynomial in standard form gives x4 9x2 4x 12 0
3 f 1x2 0 4 . Testing 1 and 1 shows x 1 is not a zero, but x 1 is. Using
synthetic division with x 1 gives
use 1 as a “divisor”
1
1
0
1
1
9
1
8
4
8
12
2 1
↓
1
1
2
1
8
2
6
12
12
0
1
↓
12
12
0
with a quotient of q1 1x2 x3 x2 8x 12 and a remainder of zero. As q1(x) is
not easily factored, we continue with synthetic division using x 2.
use 2 as a “divisor”
The result is q2 1x2 x2 x 6 with a remainder of zero.
1. The factored form is
f 1x2 1x 12 1x 22 1x2 x 62 1x 12 1x 22 2 1x 32 .
2. The graph will “cross” at x 1 and 3, and f will change sign. The graph will
“bounce”at x 2 and f will not change sign. This is illustrated in Figure 4.82
which uses closed dots since f (x) can be equal to zero.
Figure 4.82
change
3
change
2
1
no change
0
1
2
x
3. With even degree and positive lead coefficient, the end-behavior is up/up.
Working from the leftmost interval, f 1x2 7 0, the function must change sign
at x 3 (going below the x-axis), and again at x 1 (going above the
x-axis). This is supported by the y-intercept (0, 12). The graph then “bounces”
at x 2, remaining above the x-axis (no sign change). This produces the
sketch shown in Figure 4.83.
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Section 4.6 Polynomial and Rational Inequalities
Figure 4.83
End-behavior
is “up”
End-behavior
is “up”
f(0) 12
3
f(x) 0
2
1
f(x) 0
0
f(x) 0
x
f(x) 0
1
2
Figure 4.84
4. From the diagram, we see that
f 1x2 0 for x 僆 33, 14 , and at the
single point x 2. This shows the
solution for x4 4x 9x2 12 is
x 僆 3 3, 14 ´ 526. The actual graph is
shown in Figure 4.84. The graph is below
or touching the x-axis from 3 to 1
and at x 2.
20
5
5
20
Now try Exercises 19 through 24
䊳
Solving Function Inequalities Using Interval Tests
As an alternative to graphical analysis an interval test method can be used to solve
polynomial (and rational) inequalities. The x-intercepts (and vertical asymptotes in the
case of rational functions) are noted on the x-axis, then a test number is selected from
each interval. Since polynomial and rational functions are continuous over their entire
domain, the sign of the function at these test values will be the sign of the function for
all values of x in the chosen interval.
EXAMPLE 3
䊳
Solving a Polynomial Inequality
Solve the inequality x3 8 5x2 2x.
Solution
䊳
Writing the relationship in function form gives p1x2 x3 5x2 2x 8, with
solutions needed to p1x2 0. The tests for 1 and 1 show x 1 is a root, and
using 1 with synthetic division gives
use 1 as a “divisor”
1 1
↓
1
5
1
6
2
6
8
8
8
0
The quotient is q1x2 x2 6x 8, with a remainder of 0.
The factored form is p1x2 1x 12 1x2 6x 82 1x 121x 221x 42 .
The x-intercepts are (1, 0), (2, 0), and (4, 0). Plotting these intercepts creates four
intervals on the x-axis (Figure 4.85).
Figure 4.85
5 4 3 2 1
WORTHY OF NOTE
When evaluating a function using
the interval test method, it’s usually
easier to use the factored form
instead of the polynomial form, since
all you really need is whether the
result will be positive or negative.
For instance, you could likely tell
p132 13 1213 2213 42 is
going to be negative, more quickly
than p132 132 3 5132 2 2132 8.
1
0
1
2
2
3
4
5
6
3
7
8
9 x
4
Selecting a test value from each interval gives the information shown in Figure 4.86.
Figure 4.86
5 4 3 ⴚ2 1
0
x ⴚ2
x0
p(2) 24 p(0) 8
p(x) 0
p(x) 0
in 1
in 2
1
2
3
4
5
x3
p(3) 4
p(x) 0
in 3
6
7
8
9 x
x5
p(5) 18
p(x) 0
in 4
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CHAPTER 4 Polynomial and Rational Functions
The interval tests show x3 8 5x2 2x for x 僆 1q,14 ´ 3 2, 44 . The graph
is shown in Figure 4.87.
Figure 4.87
10
3
6
10
A. You’ve just seen how
we can solve polynomial
inequalities
Now try Exercises 25 and 26
䊳
B. Rational Inequalities
In general, the solution process for polynomial and rational inequalities is virtually
identical, once we recognize that vertical asymptotes also break the x-axis into intervals where function values may change sign. However, for rational functions it’s more
efficient to begin the analysis using the y-intercept or a test point, rather than endbehavior, although either will do.
EXAMPLE 4
䊳
Solving a Rational Inequality b y Analysis
Solve
Solution
䊳
x2 9
0.
x3 x2 x 1
x2 9
and we want the solution for v1x2 0.
x3 x2 x 1
The numerator and denominator are in standard form. The numerator factors easily,
and the denominator can be factored by grouping.
In function form, v1x2 1. The factored form is v1x2 1x 32 1x 32
.
1x 12 2 1x 12
2. v(x) will change sign at x 3, 3, and 1 as all have odd multiplicity, but
will not change sign at x 1 (even multiplicity). Note that zeroes of the
denominator will always be indicated by open dots (Figure 4.88) as they are
excluded from any solution set.
Figure 4.88
change
3
change
2
1
no change
0
change
2
1
3
x
3. The y-intercept is (0, 9), indicating that function values will be negative in
the interval containing zero. Working outward from this interval using the
“change/no change” approach, gives the solution indicated in Figure 4.89.
Figure 4.89
v(0) 9
change
3
v(x) 0
change
2
v(x) 0
1
no change
0
v(x) 0
1
change
2
v(x) 0
3
x
v(x) 0
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Section 4.6 Polynomial and Rational Inequalities
WORTHY OF NOTE
End-behavior can also be used to
analyze rational inequalities,
although using the y-intercept may
be more efficient. For the function
v(x) from Example 5 we have
x2
x2 9
1
3 for large
3
2
x
x x x1
x
values of x, indicating v1x2 7 0 to
the far right and v1x2 6 0 to the far
left. The analysis of each interval
can then begin from either side.
4. For v1x2 0, the solution is
x 僆 1q, 34 ´ 11, 12 ´ 11, 34 .
389
Figure 4.90
Finding a window that clearly displays all
features of rational function can sometimes be
difficult. In these cases, we could investigate
each piece separately to confirm solutions. For
this example, most of the features of v(x) can
be seen using a window size of X 僆 3 5, 54
and Y 僆 3 20, 154 , and we note the graph in
Figure 4.90 strongly tends to support our
solution.
15
5
5
20
Now try Exercises 27 through 40
䊳
If the rational inequality is not given in function form or is composed of more than
one term, start by writing the inequality with zero on one side, then combine terms into
a single expression. This is the most efficient way of determining the zeroes of the
function and the location of any vertical asymptotes.
EXAMPLE 5
䊳
Solving a Rational Inequality Using Inter val Tests
Solve the inequality
Solution
䊳
x2
1
.
x3
x3
x2
1
0. This is
x3
x3
1
x2
equivalent to v1x2 0, where v1x2 . Combining the expressions
x3
x3
on the right, we have
Rewrite the inequality with zero on one side:
v1x2 1x 22 1x 32 11x 32
LCD is 1x 321x 32
1x 32 1x 32
2
x x6x3
1x 32 1x 32
x2 3
1x 32 1x 32
multiply
simplify
1x 132 1x 132
. x2 k 1x 1k21x 1k2
1x 32 1x 32
2. The zeroes and asymptotes will occur at x 13, 13, 3, and 3, which we
plot on the x-axis, creating five intervals (Figure 4.91).
1. The factored form is v1x2 Figure 4.91
3
2
1
0
1
3. From left to right, we select one test value
from each interval created by the zeroes
and vertical asymptotes. We chose:
x 4, 2, 0, 2, and 4 (see Figure 4.92).
The results are shown in Figure 4.93, with
the conclusion noted beneath each interval.
2
3
x
Figure 4.92
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CHAPTER 4 Polynomial and Rational Functions
Figure 4.93
v(0) a
3
3
2
3
1
v(x) 0 v(x) 0
0
v(x) 0
x
2
3
v(x) 0 v(x) 0
1
x2
1
is
x3
x3
x 僆 13, 134 ´ 3 13, 32 .
Figure 4.94
4. The solution for
To check these solutions graphically,
1
we subtract
from both sides and graph
x3
x2
1
Y1 to look for intervals where
x3
x3
the graph is below the x-axis. The graph is shown in Figure 4.94 and verifies our
solution.
B. You’ve just seen how we
can solve rational inequalities
Now try Exercises 41 through 60
䊳
C. Applications of Inequalities
Applications of inequalities come in many varieties. In addition to stating the solution
algebraically, these exercises often compel us to consider the context of each application as we state the solution set.
EXAMPLE 6
䊳
Solving Applications of Inequalities
The velocity of a particle (in feet per second) as it floats through air turbulence is
given by V1t2 t5 10t4 35t3 50t2 24t, where t is the time in seconds and
0 6 t 6 4.5. During what intervals of time is the particle moving in the positive
direction 3V1t2 7 0 4 ?
Solution
䊳
Begin by writing V in factored form. Testing 1 and 1 shows t 1 is a root.
Factoring out t gives V1t2 t1t4 10t3 35t2 50t 24), and using t 1 with
synthetic division yields
use 1 as a “divisor”
1 1
↓
1
10
1
9
35
9
26
50
26
24
24
24
0
The quotient is q1 1t2 t3 9t2 26t 24. Using t 2, we continue with the
division on q1(t) which gives:
use 2 as a “divisor”
2
1
↓
1
9
2
7
This shows V1t2 t1t 12 1t 22 1t2 7t 122 .
26
14
12
24
24
0
1. The completely factored form is V1t2 t1t 12 1t 22 1t 32 1t 42.
2. All zeroes have odd multiplicity and function values will change sign.
3. With odd degree and a positive leading coefficient, end-behavior is down/up.
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Section 4.6 Polynomial and Rational Inequalities
Function values will be negative in the far left interval and alternate in sign
thereafter. The solution diagram is shown in the figure (the parentheses shown on
the number line indicate the domain given).
Since end-behavior is down/up, function values
are negative in this interval, and will alternate thereafter.
change
1
V(t) 0
)
0
change
change
change
change
1
2
3
4
V(t) 0 V(t) 0 V(t) 0 V(t) 0
)
t
V(t) 0
4. For V1t2 7 0, the solution is t 僆 10, 12 ´ 12, 32 ´ 14, 4.52 . The particle is
moving in the positive direction in these time intervals.
Now try Exercises 63 through 70
䊳
Figure 4.95
5
C. You’ve just seen how
we can solve applications of
inequalities
To verify the analysis in Example 6, we graph
V(t) using the window X 僆 3 1, 54 and
Y 僆 3 5, 54 . As Figure 4.95 shows, function
values are positive (graph is above the x-axis)
when t 僆 10, 12 ´ 12, 32 ´ 14, 4.52 .
1
5
5
4.6 EXERCISES
䊳
CONCEPTS AND VOCABULAR Y
Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.
1. To solve a polynomial or rational inequality, begin
by plotting the location of all zeroes and
asymptotes (if they exist), then consider the
of each.
2. For strict inequalities, the zeroes are
from the solution set. For nonstrict inequalities,
zeroes are
. The values at which vertical
asymptotes occur are always
.
3. Since the graph of g1x2 x4 3, opens upward
with a vertex at (0, 3) the solution set for g1x2 7 0
is
.
4. To solve a polynomial/rational inequality, it helps
to find the sign of f in some interval. This can
quickly be done using the
or
of the function.
5. Compare/Contrast the process for solving
1
x2 3x 4 0 with 2
0. Are there
x 3x 4
similarities? What are the differences?
6. Compare/Contrast the process for solving
1x 12 1x 32 1x2 12 7 0 with
1x 12 1x 32 7 0. Are there similarities? What
are the differences?
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䊳
4–86
CHAPTER 4 Polynomial and Rational Functions
DEVELOPING YOUR SKILLS
39.
x2 5x 14
7 0
x3 x2 5x 3
40.
x2 2x 8
6 0
x3 5x2 3x 9
11. 1x 22 3 1x 22 2 1x 42 0
41.
1
2
x
x2
42.
3
5
x
x3
13. x 4x 1 6 0
43.
x3
1
7
x 17
x1
44.
x2
1
6
x5
x7
45.
x1
x2
x2
x3
46.
x3
x1
x6
x4
47.
x2
7 0
x2 9
48.
x2 4
6 0
x3
49.
x3 1
7 0
x2 1
50.
x2 4
6 0
x3 8
51.
x4 5x2 36
7 0
x2 2x 1
52.
x4 3x2 4
6 0
x2 x 20
Solve each polynomial inequality indicated using a
number line and the behavior of the graph at each zero.
Write all answers in interval notation.
7. 1x 32 1x 52 6 0
8. 1x 22 1x 72 6 0
9. 1x 12 2 1x 42 0 10. 1x 62 1x 12 2 0
12. 1x 12 3 1x 22 2 1x 32 0
2
14. x 6x 4 7 0
2
15. x x 5x 3 0
3
2
16. x x 8x 12 0
3
2
17. x3 7x 6 7 0
18. x3 13x 12 7 0
19. x4 10x2 7 9
20. x4 36 6 13x2
21. x4 9x2 7 4x 12
22. x4 16 7 5x3 20x
23. x4 6x3 8x2 6x 9
24. x4 3x2 8 4x3 10x
Solve each inequality using the interval test method.
25. 4x 12 6 x 3x
3
26. x3 8 6 5x2 2x
28.
2
27.
x2 4x 21
6 0
x3
x2 x 6
0
x2 1
53. x2 2x 15
54. x2 3x 18
55. x3 9x
56. x3 4x
Match the correct solution with the inequality and
graph given.
57. R1x2 0
y
5
Solve each rational inequality indicated using a number
line and the behavior of the graph at each zero. Write
all answers in interval notation.
x3
29. f 1x2 ; f 1x2 0
x2
x4
30. F1x2 ; F1x2 0
x1
x1
31. g1x2 2
; g1x2 6 0
x 4x 4
32. G1x2 33.
x3
; G1x2 7 0
x 2x 1
2
2x
0
2
x x6
2x x2
35. 2
6 0
x 4x 5
37.
x2 4
0
x3 13x 12
34.
1x
0
2
x 2x 8
x2 3x
36. 2
7 0
x 2x 3
38.
x2 x 6
0
x3 7x 6
y R(x)
5
5 x
5
a.
b.
c.
d.
e.
x 僆 1q, 12 ´ 10, 22
x 僆 30, 1 4 ´ 12, q 2
x 僆 35, 14 ´ 32, 5 4
x 僆 1q, 12 ´ 3 0, 22
none of these
58. g1x2 0
a. x 僆 14, 0.52 ´ 14, q 2
b. x 僆 30.5, 44 ´ 3 4, 54
c. x 僆 1q, 42 ´ 10.5, 42 5
d. x 僆 34, 0.54 ´ 34, q 2
e. none of these
y
5
y g(x)
5 x
5
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Precalculus—
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59. f 1x2 6 0
y
5
5
y f(x)
60. r1x2 0
a.
b.
c.
d.
e.
x 僆 15, 22 ´ 13, 52
x 僆 1q, 22 ´ 12, 12 ´ 13, q 2
x 僆 1q, 22 ´ 13, q2
x 僆 1q, 22 ´ 12, 14 ´ 33, q 2
none of these
y
5
5
5 x
y r(x)
5 x
5
5
䊳
393
Section 4.6 Polynomial and Rational Inequalities
a.
b.
c.
d.
e.
x 僆 1q, 22 ´ 31, 14 ´ 33, q 2
x 僆 12, 14 ´ 31, 22 ´ 12, 34
x 僆 1q, 22 ´ 12, q 2
x 僆 12, 12 ´ 11, 22 ´ 12, 34
none of these
WORKING WITH FORMULAS
61. Discriminant of the reduced cubic x3 ⴙ px ⴙ q ⴝ 0: D ⴝ ⴚ14p3 ⴙ 27q2 2
The discriminant of a cubic equation is less well known than that of the quadratic, but serves the same purpose.
The discriminant of the reduced cubic is given by the formula shown, where p is the linear coefficient and q is
the constant term. If D 7 0, there will be three real and distinct roots. If D 0, there are still three real roots, but
one is a repeated root (multiplicity two). If D 6 0, there are one real and two complex roots. Suppose we wish to
study the family of cubic equations where q p 1.
a. Verify the resulting discriminant is D 14p3 27p2 54p 272.
b. Determine the values of p and q for which this family of equations has a repeated real root. In other words,
solve the equation 14p3 27p2 54p 272 0 using the rational zeroes theorem and synthetic division
to write D in completely factored form.
c. Use the factored form from part (b) to determine the values of p and q for which this family of equations has
three real and distinct roots. In other words, solve D 7 0.
d. Verify the results of parts (b) and (c) on a graphing calculator.
62. Coordinates for the folium of Descartes:
The interesting relation shown here is called the folium (leaf) of Descartes. The folium is most
3kx
often graphed using what are called parametric equations, in which the coordinates a and b are
a
1 x3
expressed in terms of the parameter x (“k” is a constant that affects the size of the leaf). Since
μ
each is an individual function, the x- and y-coordinates can be investigated individually in
3kx2
2
b
3x
3x
rectangular coordinates using F1x2 and G1x2 (assume k 1 for now).
1 x3
3
3
1x
1x
a. Graph each function using the techniques from this chapter.
b
b. Will F(x) ever be equal to G(x)? If so, for what values of x?
(a, b)
c. According to your graph, for what values of x will the x-coordinate of
3x
a
7 0.
the folium be positive? In other words, solve F1x2 1 x3
d. For what values of x will the y-coordinate of the folium be positive? Solve
3x2
7 0.
G1x2 Folium of Descartes
1 x3
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䊳
CHAPTER 4 Polynomial and Rational Functions
4–88
APPLICATIONS
Deflection of a beam: The amount of
deflection in a rectangular wooden beam
of length L ft can be approximated by
d1x2 k1x3 3L2x 2L3 2, where k is a
constant that depends on the
characteristics of the wood and the force
applied, and x is the distance from the
unsupported end of the beam 1x 6 L2.
Weight
Deflection
63. Find the equation for a beam 8 ft long and use it
for the following:
d1x2
a. For what distances x is the quantity
less
k
than 189 units?
b. What is the amount of deflection 4 ft from the
unsupported end (x 4)?
d1x2
c. For what distances x is the quantity
k
greater than 475 units?
d. If safety concerns prohibit a deflection of more
than 648k units, what is the shortest distance
from the end of the beam that the force can be
applied?
64. Find the equation for a beam 9 ft long and use it
for the following:
d1x2
a. For what distances x is the quantity
less
k
than 216 units?
b. What is the amount of deflection 4 ft from the
unsupported end (x 4)?
d1x2
c. For what distances x is the quantity
k
greater than 550k units?
d. Compare the answer to 63b with the answer to
64b. What can you conclude?
Average speed for a
round-trip: Surprisingly,
the average speed of a
round-trip is not the sum
of the average speed in
each direction divided by
two. For a fixed distance
D, consider rate r1 in time
t1 for one direction, and
rate r2 in time t2 for
D
D
the other, giving r1 and r2 . The average speed for
t1
t2
2D
the round-trip is R .
t1 t2
65. The distance from St. Louis, Missouri, to
Springfield, Illinois, is approximately 80 mi.
Suppose that Sione, due to the age of his vehicle,
made the round-trip with an average speed of
40 mph.
a. Use the relationships stated to verify that
20r1
r2 .
r1 20
b. Discuss the meaning of the horizontal and
vertical asymptotes in this context.
c. Verify algebraically the speed returning would
be greater than the speed going for
20 6 r1 6 40. In other words, solve the
20r1
7 r1 using the ideas from
inequality
r1 20
this section.
66. The distance from Boston, Massachusetts, to
Hartford, Connecticut, is approximately 100 mi.
Suppose that Stella, due to excellent driving
conditions, made the round-trip with an average
speed of 60 mph.
a. Use the relationships above to verify that
30r1
r2 .
r1 30
b. Discuss the meaning of the horizontal and
vertical asymptotes in this context.
c. Verify algebraically the speed returning would
be greater than the speed going for
30 6 r1 6 60. In other words, solve the
30r1
7 r1 using the ideas from
inequality
r1 30
this section.
Electrical resistance and temperature: The amount of
electrical resistance R in a medium depends on the
temperature, and for certain materials can be modeled by
the equation R1t2 0.01t 2 0.1t k, where R(t) is the
resistance (in ohms ) at temperature t 1t 0°2 in degrees
Celsius, and k is the resistance at t 0°C.
67. Suppose k 30 for a certain medium. Write the
resistance equation and use it to answer the
following.
a. For what temperatures is the resistance less
than 42 ?
b. For what temperatures is the resistance greater
than 36 ?
c. If it becomes uneconomical to run electricity
through the medium for resistances greater
than 60 , for what temperatures should the
electricity generator be shut down?
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Section 4.6 Polynomial and Rational Inequalities
68. Suppose k 20. Write the resistance equation and
solve the following.
a. For what temperatures is the resistance less
than 26 ?
b. For what temperatures is the resistance greater
than 40 ?
c. If it becomes uneconomical to run electricity
through the medium for resistances greater
than 50 , for what temperatures should the
electricity generator be shut down?
69. Sum of consecutive squares: The sum of the first
n squares 12 22 32 p n2 is given by the
2n3 3n2 n
. Use the equation
formula S1n2 6
to solve the following inequalities.
a. For what number of consecutive squares is
S1n2 30?
䊳
395
b. For what number of consecutive squares is
S1n2 285?
c. What is the maximum number of consecutive
squares that can be summed without the result
exceeding three digits?
70. Sum of consecutive cubes: The sum of the first n
cubes 13 23 33 p n3 is given by the
n4 2n3 n2
. Use the equation
formula S1n2 4
to solve the following inequalities.
a. For what number of consecutive cubes is
S1n2 100?
b. For what number of consecutive cubes is
S1n2 784?
c. What is the maximum number of consecutive
cubes that can be summed without the result
exceeding three digits?
EXTENDING THE CONCEPT
71. (a) Is it possible for the solution set of a
polynomial inequality to be all real numbers? If
not, discuss why. If so, provide an example. (b) Is
it possible for the solution set of a rational
inequality to be all real numbers? If not, discuss
why. If so, provide an example.
72. As in our earlier studies, if n is an even number, the
n
expression 1 A represents a real number only if
A 0. Use this idea to find the domain of the
following functions.
a. f 1x2 22x3 x2 16x 15
4
b. g1x2 22x3 x2 22x 24
x2
4
c. p1x2 B x2 2x 35
x2 1
d. q1x2 B x2 x 6
74. Using the tools of calculus, it can be shown that
f 1x2 x4 4x3 12x2 32x 39 is increasing
in the intervals where F1x2 x3 3x2 6x 8
is positive. Solve the inequality F1x2 7 0 using the
ideas from this section, then verify f 1x2c in these
intervals by graphing f on a graphing calculator and
using the TRACE feature.
75. Using the tools of calculus, it can be shown that
x2 3x 4
is decreasing in the intervals
r 1x2 x8
x2 16x 28
where R1x2 is negative. Solve
1x 82 2
the inequality R1x2 6 0 using the ideas from this
section, then verify r 1x2T in these intervals by
graphing r on a graphing calculator and using the
TRACE
feature.
73. Find one polynomial inequality and one rational
inequality that have the solution
x 僆 1q, 22 ´ 10, 12 ´ 11, q 2.
䊳
MAINTAINING YOUR SKILLS
Exercise 76
76. (2.2) Use the graph of f (x)
given to sketch the graph of
y f 1x 22 3.
y
5
is a removable discontinuity, repair the break using
an appropriate piecewise-defined function.
y f(x)
77. (2.5/4.5) Graph the function
f 1x2 x2 2x 8
. If there
x4
5
5 x
5
78. (Appendix A.6) Solve the equation
x
1
116 x 2.
2
2
Check solutions in the original equation.
79. (2.3) Solve the absolute value inequality:
2x 3 5 7
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CHAPTER 4 Polynomial and Rational Functions
MAKING CONNECTIONS
Making Connections: Graphically , Symbolically , Numerically, and V erbally
Eight graphs (a) through (h) are given. Match the characteristics shown in 1 through 16 to one of the eight graphs.
y
(a)
54321
1
2
3
4
5
1. ____
1 2 3 4 5 x
5
4
3
2
1
54321
1
2
3
4
5
1 2 3 4 5 x
1
1x 42 1x 12 1x 32
6
2. ____ y 1x 22 2 4
y
(c)
5
4
3
2
1
54321
1
2
3
4
5
1 2 3 4 5 x
y
(f)
5
4
3
2
1
54321
1
2
3
4
5
5
4
3
2
1
54321
1
2
3
4
5
1 2 3 4 5 x
y
(e)
y
(b)
5
4
3
2
1
5
4
3
2
1
54321
1
2
3
4
5
1 2 3 4 5 x
5
4
3
2
1
54321
1
2
3
4
5
1 2 3 4 5 x
y
(g)
y
(d)
1 2 3 4 5 x
y
(h)
5
4
3
2
1
54321
1
2
3
4
5
1 2 3 4 5 x
1
9. ____ 1y 42 1x 42
6
3
10. ____ y x
2
3. ____ min degree 3, y-int: (0, 1)
11. ____ axis of symmetry x 2, f 112 3
4. ____ min degree 4, one repeated root,
relative max y 0 at x 1
12. ____ min degree 4, one repeated root,
lead coefficient a 6 0
5. ____ y x1x 42
13. ____ axis of symmetry x 2, f 132 3
7. ____ f 112 4, f 122 1
15. ____ f 1x2c only for x 僆 13, 12 ´ 11, q 2
6. ____ f 1x2 7 0 only for x 僆 14, 02
8. ____ f 1x2c only for x 僆 12, 22
14. ____ f 1x2 7 0 for x 僆 14, 12 ´ 13, q 2
16. ____ f 1x2 4.2 for x 2 and x 4
SUMMAR Y AND CONCEPT REVIEW
SECTION 4.1
Synthetic Division; the Remainder and Factor Theor ems
KEY CONCEPTS
• Synthetic division is an abbreviated form of long division. Only the coefficients of the dividend are used, since
“standard form” ensures like place values are aligned. Zero placeholders are used for “missing” terms. The
“divisor” must be linear with leading coefficient 1.
• To divide a polynomial by x c, use c in the synthetic division; to divide by x c, use c.
• After setting up the synthetic division template, drop the leading coefficient of the dividend into place, then
multiply in the diagonal direction, place the product in the next column, and add in the vertical direction,
continuing to the last column.
• The final sum is the remainder r, the numbers preceding it are the coefficients of q(x).
• Remainder theorem: If p(x) is divided by x c, the remainder is equal to p(c). The theorem can be used to
evaluate polynomials at x c.
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Summary and Concept Review
397
• Factor theorem: If p1c2 0, then c is a zero of p and 1x c2 is a factor. Conversely, if 1x c2 is a factor of p,
then p1c2 0. The theorem can be used to factor a polynomial or build a polynomial from its zeroes.
• The remainder and factor theorems also apply when c is a complex number.
EXERCISES
Divide using long division and clearly identify the quotient and remainder:
x3 4x2 5x 6
x3 2x 4
1.
2.
x2
x2 x
3. Use the factor theorem to show that x 7 is a factor of 2x4 13x3 6x2 9x 14.
h1x2
x3 4x 5
.
4. Complete the division and write h(x) as h1x2 d1x2q1x2 r1x2 , given
x2
d1x2
5. Use the factor theorem to help factor p1x2 x3 2x2 11x 12 completely.
6. Use the factor and remainder theorems to factor h, given x 4 is a zero: h1x2 x4 3x3 4x2 2x 8.
Use the remainder theorem:
7. Show x 12 is a zero of V: V1x2 4x3 8x2 3x 1.
8. Show x 3i is a zero of W: W1x2 x3 2x2 9x 18.
9. Find h172 given h1x2 x3 9x2 13x 10.
Use the factor theorem:
10. Find a degree 3 polynomial in standard form with zeroes x 1, x 15, and x 15.
11. Find a fourth-degree polynomial in standard form with one real zero, given x 1 and x 2i are zeroes.
12. Use synthetic division and the remainder theorem to answer: At a busy shopping mall, customers are constantly
coming and going. One summer afternoon during the hours from 12 o’clock noon to 6 in the evening, the number
of customers in the mall could be modeled by C1t2 3t3 28t2 66t 35, where C(t) is the number of
customers (in tens), t hours after 12 noon. (a) How many customers were in the mall at noon? (b) Were more
customers in the mall at 2:00 or at 3:00 P.M.? How many more? (c) Was the mall busier at 1:00 P.M. (after lunch)
or 6:00 P.M. (around dinner time)?
SECTION 4.2
The Zer oes of Polynomial Functions
KEY CONCEPTS
• Fundamental theorem of algebra: Every complex polynomial of degree n 1 has at least one complex zero.
• Linear factorization theorem: Every complex polynomial of degree n 1 has exactly n linear factors, and can be
written in the form p1x2 a1x c1 2 1x c2 2 p 1x cn 2, where a 0 and c1, c2, p , cn are (not necessarily
distinct) complex numbers.
• For a polynomial p in factored form with repeated factors 1x c2 m, c is a zero of multiplicity m. If m is odd, c is a
zero of odd multiplicity; if m is even, c is a zero of even multiplicity.
• Corollaries to the linear factorization theorem:
I. If p is a polynomial with real coefficients, p can be factored into linear factors (not necessarily distinct) and
irreducible quadratic factors having real coefficients.
II. If p is a polynomial with real coefficients, the complex zeroes of p must occur in conjugate pairs. If
a bi 1b 02, is a zero, then a bi is also a zero.
III. If p is a polynomial with degree n 1, then p will have exactly n zeroes (real or complex), where zeroes of
multiplicity m are counted m times.
• Intermediate value theorem: If p is a polynomial with real coefficients where p(a) and p(b) have opposite signs,
then there is at least one c between a and b such that p1c2 0.
p
• Rational zeroes theorem: If a real polynomial has integer coefficients, rational zeroes must be of the form q, where
p is a factor of the constant term and q is a factor of the leading coefficient.
• Descartes’ rule of signs, upper and lower bounds property, tests for 1 and 1, and graphing technology can all be
used with the rational zeroes theorem to factor, solve, and graph polynomial functions.
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CHAPTER 4 Polynomial and Rational Functions
EXERCISES
Using the tools from this section,
13. List all possible rational zeroes of
14. Find all rational zeroes of
p1x2 4x3 16x2 11x 10.
p1x2 4x3 16x2 11x 10.
15. Write P1x2 2x3 3x2 17x 12
16. Prove that h1x2 x4 7x2 2x 3
in completely factored form.
has no rational zeroes.
17. Identify two intervals (of those given) that contain a zero of P1x2 x4 3x3 8x2 12x 6: 32, 1 4,
[1, 2], [2, 3], [4, 5]. Then verify your answer using a graphing calculator.
18. Discuss the number of possible positive, negative, and complex zeroes for g1x2 x4 3x3 2x2 x 30.
Then identify which combination is correct using a graphing calculator.
SECTION 4.3
Graphing Polynomial Functions
KEY CONCEPTS
• All polynomial graphs are smooth, continuous curves.
• A polynomial of degree n has at most n 1 turning points. The precise location of these turning points are the
local maximums or local minimums of the function.
• If the degree of a polynomial is odd, the ends of its graph will point in opposite directions (like y mx2. If the
degree is even, the ends will point in the same direction (like y ax2 2. The sign of the lead coefficient determines
the actual behavior.
• The “behavior” of a polynomial graph near its zeroes is determined by the multiplicity of the zero. For any factor
1x c2 m, the graph will “cross through” the x-axis if m is odd and “bounce off” the x-axis (touching at just one
point) if m is even. The larger the value of m, the flatter (more compressed) the graph will be near c.
• To “round-out” a graph, additional midinterval points can be found between known zeroes.
• These ideas help to establish the Guidelines for Graphing Polynomial Functions. See page 345.
EXERCISES
State the degree, end-behavior, and y-intercept, but do not graph.
19. f 1x2 3x5 2x4 9x 4
20. g1x2 1x 12 1x 22 2 1x 22
Graph using the Guidelines for Graphing Polynomials.
21. p1x2 1x 12 3 1x 22 2
22. q1x2 2x3 3x2 9x 10
23. h1x2 x4 6x3 8x2 6x 9
24. For the graph of P(x) shown, (a) state whether the degree of P is even or odd, (b) use the
graph to locate the zeroes of P and state whether their multiplicity is even or odd, and
(c) find the minimum possible degree of P and write it in factored form. Assume all zeroes
are real.
SECTION 4.4
y
5
4
3
2
1
54321
1
2
3
4
5
Graphing Rational Functions
KEY CONCEPTS
• A rational function is one of the form V1x2 •
•
•
•
•
p1x2
, where p and d are polynomials and d1x2 0.
d1x2
The domain of V is all real numbers, except the zeroes of d.
If zero is in the domain of V, substitute 0 for x to find the y-intercept.
The zeroes of V (if they exist), are solutions to p1x2 0.
The line y k is a horizontal asymptote of V if as 冟x冟 increases without bound, V(x) approaches k.
p1x2
If
is in simplest form, vertical asymptotes will occur at the zeroes of d.
d1x2
1 2 3 4 5 x
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Summary and Concept Review
• The line x h is a vertical asymptote of V if as x approaches h, V(x) increases/decreases without bound.
• If the degree of p is less than the degree of d, y 0 (the x-axis) is a horizontal asymptote. If the degree of p is
equal to the degree of d, y a
is a horizontal asymptote, where a is the leading coefficient of p, and b is the
b
leading coefficient of d.
• The Guidelines for Graphing Rational Functions can be found on page 361.
EXERCISES
x2 9
, state the following but do not graph: (a) domain (in set notation),
x2 3x 4
(b) equations of the horizontal and vertical asymptotes, (c) the x- and y-intercept(s), and (d) the value of V(1).
1x 12 2
, will the function change sign at x 1? Will the function change sign at x 2? Justify
26. For v1x2 x2
your responses.
25. For the function V1x2 Graph using the Guidelines for Graphing Rational Functions.
x2 4x
2x2
27. v1x2 2
28. t1x2 2
x 4
x 5
29. Use the vertical asymptotes, x-intercepts, and their multiplicities to construct an equation
that corresponds to the given graph. Be sure the y-intercept on the graph matches the value
given by your equation. Assume these features are integer-valued. Check your work on a
graphing calculator.
y
5
4
3
2
1
108642
1
2 4 6 8 10 x
2
30. The average cost of producing a popular board game is given by the function
3
5000 15x
4
5
A1x2 ; x 1000. (a) Identify the horizontal asymptote of the function and
x
explain its meaning in this context. (b) To be profitable, management believes the average cost must be below
$17.50. What levels of production will make the company profitable?
SECTION 4.5
Additional Insights into Rational Functions
KEY CONCEPTS
p1x2
is not in simplest form, with p and d sharing factors of the form x c, the graph will have a
• If V d1x2
removable discontinuity (a hole or gap) at x c. The discontinuity can be “removed” (repaired) by redefining V
using a piecewise-defined function.
p1x2
is in simplest form, and the degree of p is greater than the degree of d, the graph will have an oblique
• If V d1x2
or nonlinear asymptote, as determined by the quotient polynomial after division. If the degree of p is greater by 1,
the result is a slant (oblique) asymptote. If the degree of p is greater by 2, the result is a parabolic asymptote.
EXERCISES
31. Determine if the graph of h will have a vertical asymptote or a removable discontinuity, then graph the function
x3 2x2 9x 18
h1x2 .
x2
x2 3x 4
32. Sketch the graph of h1x2 . If there is a removable discontinuity, repair the break by redefining h
x1
using an appropriate piecewise-defined function.
Graph the functions using the Guidelines for Graphing Rational Functions.
x3 7x 6
x2 2x
33. h1x2 34. t1x2 x3
x2
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CHAPTER 4 Polynomial and Rational Functions
35. The cost to make x thousand party favors is given by C1x2 x2 2x 6, where x 1 and C is in thousands
x2 2x 6
of dollars. For the average cost of production A1x2 , (a) graph the function, (b) use the graph to
x
estimate the level of production that will make average cost a minimum, and (c) state the average cost of a single
party favor at this level of production.
SECTION 4.6
Polynomial and Rational Inequalities
KEY CONCEPTS
• To solve polynomial inequalities, write P(x) in factored form and note the multiplicity of each real zero.
• Plot real zeroes on a number line. The graph will cross the x-axis at zeroes of odd multiplicity (P will change
sign), and bounce off the axis at zeroes of even multiplicity (P will not change sign).
• Use the end-behavior, y-intercept, or a test point to determine the sign of P in a given interval, then label all other
intervals as P1x2 7 0 or P1x2 6 0 by analyzing the multiplicity of neighboring zeroes. Use the resulting diagram
to state the solution.
• The solution process for rational inequalities and polynomial inequalities is virtually identical, considering that
vertical asymptotes also create intervals where function values may change sign, depending on their multiplicity.
• Polynomial and rational inequalities can also be solved using an interval test method. Since polynomials and
rational functions are continuous on their domains, the sign of the function at any one point in an interval will
be the same as for all other points in that interval.
EXERCISES
Solve each inequality indicated using a number line and the behavior of the graph at each zero.
x
1
x2 3x 10
36. x3 x2 7 10x 8
37.
38.
0
x
x2
x2
PRACTICE TEST
1. Complete the square to write each function as a
transformation. Then graph each function and label
the vertex and all intercepts (if they exist).
a. f 1x2 x2 10x 16
1
b. g1x2 x2 4x 16
2
2. The graph of a quadratic function has a vertex of
11, 22 , and passes through the origin. Find the
other intercept, and the equation of the graph in
standard form.
3. Suppose the function d1t2 t2 14t models the
depth of a scuba diver at time t, as she dives
underwater from a steep shoreline, reaches a
certain depth, and swims back to the surface.
a. What is her depth after 4 sec? After 6 sec?
b. What was the maximum depth of the dive?
c. How many seconds was the diver beneath the
surface?
4. Compute the quotient using long division:
x3 3x2 5x 2
.
x2 2x 1
5. Find the quotient and remainder using synthetic
x3 4x2 5x 20
division:
.
x2
6. Use the remainder theorem to show 1x 32 is a
factor of x4 15x2 10x 24.
7. Given f 1x2 2x3 4x2 5x 2, find the value of
f 132 using synthetic division and the remainder
theorem.
8. Given x 2 and x 3i are two zeroes of a real
polynomial P1x2 with degree 3. Use the factor
theorem to find P1x2 .
9. Factor the polynomial and state the multiplicity of each
zero: Q1x2 1x2 3x 22 1x3 2x2 x 22.
10. Given C1x2 x4 x3 7x2 9x 18, (a) use the
rational zeroes theorem to list all possible rational
zeroes; (b) apply Descartes’ rule of signs to count
the number of possible positive, negative, and
complex zeroes; and (c) use this information along
with the tests for 1 and 1, synthetic division, and
the factor theorem to factor C completely.
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Calculator Exploration and Discovery
11. Over a 10-yr period, the balance of payments (deficit
versus surplus) for a small county was modeled by
the function f 1x2 12x3 7x2 28x 32, where
x 0 corresponds to 2000 and f(x) is the deficit or
surplus in millions of dollars. (a) Use the rational
roots theorem and synthetic division to find the years
the county “broke even” (debt surplus 0 ) from
2000 to 2010. (b) How many years did the county
run a surplus during this period? (c) What was the
surplus/deficit in 2007?
12. Sketch the graph of f 1x2 1x 32 1x 12 3 1x 22 2
using the degree, end-behavior, x- and y-intercepts,
zeroes of multiplicity, and a few “midinterval” points.
13. Use the Guidelines for Graphing Polynomials to
graph g1x2 x4 9x2 4x 12.
14. Use the Guidelines for Graphing Rational Functions
x2
to graph h1x2 2
.
x 3x 4
15. Suppose the cost of cleaning contaminated soil from
300x
, where
a dump site is modeled by C1x2 100 x
C(x) is the cost (in $1000s) to remove x% of the
contaminants. Graph using x 僆 30, 100 4, and use
the graph to answer the following questions.
a. What is the significance of the vertical asymptote
(what does it mean in this context)?
b. If EPA regulations are changed so that 85% of the
contaminants must be removed, instead of the
80% previously required, how much additional
cost will the new regulations add? Compare the
cost of the 5% increase from 80% to 85% with
the cost of the 5% increase from 90% to 95%.
What do you notice?
c. What percent of the pollutants can be removed if
the company budgets $2,200,000?
401
17. Find the level of production that will minimize the
average cost of an item, if production costs are
modeled by C1x2 2x2 25x 128, where C1x2 is
the cost to manufacture x hundred items.
18. Solve each inequality
a. x3 13x 12
b.
2
3
6
x
x2
19. Suppose the concentration of a chemical in the
bloodstream of a large animal h hr after injection
into muscle tissue is modeled by the formula
2h2 5h
.
C1h2 3
h 55
a. Sketch a graph of the function for the intervals
x 僆 35, 20 4, y 僆 3 0, 14 .
b. Where is the vertical asymptote? Does it play a
role in this context?
c. What is the concentration after 2 hr? After 8 hr?
d. How long does it take the concentration to fall
below 20% 3C1h2 6 0.2 4 ?
e. When does the maximum concentration of the
chemical occur? What is this maximum?
f. Describe the significance of the horizontal
asymptote in this context.
20. Use the vertical asymptotes, x-intercepts, and
their multiplicities to construct an equation that
corresponds to the given graph. Be sure the
y-intercept on the graph matches the value given
by your equation. Assume these features are
integer-valued. Check your work on a graphing
calculator.
16. Graph using the Guidelines for Graphing Rational
Functions.
x3 x2 9x 9
a. r1x2 x2
x3 7x 6
b. R1x2 x2 4
y
8
6
8x
8
CALCULATOR EXPLORA TION AND DISCOVER Y
Complex Zer oes, Repeated Zer oes, and Inequalities
This Calculator Exploration and Discovery will explore the relationship between the solution of a polynomial (or
rational) inequality and the complex zeroes and repeated zeroes of the related function. After all, if complex zeroes
can never create an x-intercept, how do they affect the function? And if a zero of even multiplicity never crosses the
x-axis (always bounces), can it still affect a nonstrict (less than or equal to or greater than or equal to) inequality?
These are interesting and important questions, with numerous avenues of exploration. To begin, consider the function
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CHAPTER 4 Polynomial and Rational Functions
Y1 1x 32 2 1x3 12. In completely factored form Y1 1x 32 2 1x 12 1x2 x 12. This is a polynomial
function of degree 5 with two real zeroes (one repeated), two complex zeroes (the quadratic factor is irreducible), and
after viewing the graph on Figure 4.99, four turning points. From the graph (or by analysis), we have Y1 0 for
x 1. Now let’s consider Y2 1x 32 2 1x 12, the same function as Y1, less the quadratic factor. Since complex
zeroes never “cross the x-axis” anyway, the removal of this factor cannot affect the solution set of the inequality! But
how does it affect the function? Y2 is now a function of degree three, with three real zeroes (one repeated) and only
two turning points (Figure 4.100). But even so, the solution to Y2 0 is the same as for Y1 0: x 1. Finally, let’s
look at Y3 x 1, the same function as Y2 but with the repeated zero removed. The key here is to notice that since
1x 32 2 will be nonnegative for any value of x, it too does not change the solution set of the “less than or equal to
inequality,” only the shape of the graph. Y3 is a function of degree 1, with one real zero and no turning points, but the
solution interval for Y3 0 is the same solution interval as Y2 and Y1: x 1 (see Figure 4.101).
Figure 4.99
Figure 4.100
6
Figure 4.101
6
6
5
5
5
5
13
13
5
5
13
Explore these relationships further using the exercises in (A) and (B) given, using a “greater than or equal to”
inequality. Begin by writing Y1 in completely factored form.
(A) Y1 1x3 6x2 322 1x2 12
Y2 x3 6x2 32
Y3 x 2
(B) Y1 1x 32 2 1x3 2x2 x 22
Y2 1x 32 2 1x 22
Y3 x 2
Exercise 1: Based on what you’ve noticed, comment on how the irreducible quadratic factors of a polynomial affect
its graph. What role do they play in the solution of inequalities?
Exercise 2: How do zeroes of even multiplicity affect the solution set of nonstrict inequalities (less/greater than or
equal to)?
STRENGTHENING CORE SKILLS
Solving Inequalities Using the Push Principle
The most common method for solving polynomial inequalities involves finding the zeroes of the function and checking
the sign of the function in the intervals between these zeroes. In Section 4.6, we relied on the end-behavior of the graph,
the sign of the function at the y-intercept, and the multiplicity of the zeroes to determine the solution. There is a third
method that is more conceptual in nature, but in many cases highly efficient. It is based on two very simple ideas, the
first involving only order relations and the number line:
A. Given any number x and constant k 7 0: x 7 x k and x 6 x k.
x4
x4x
x
x3
x
xx3
This statement simply reinforces the idea that if a is left of b on the number line, then a 6 b. As shown in the
diagram, x 4 6 x and x 6 x 3, from which x 4 6 x 3 for any x.
B. The second idea reiterates well-known ideas regarding the multiplication of signed numbers. For any number of factors:
if there is an even number of negative factors, the result is positive;
if there is an odd number of negative factors, the result is negative.
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Cumulative Review Chapters 1–4
403
These two ideas work together to solve inequalities using what we’ll call the push principle. Consider the inequality
x2 x 12 7 0. The factored form is 1x 42 1x 32 7 0 and we want the product of these two factors to be positive.
From (A), both factors will be positive if 1x 42 is positive, since it’s the smaller of the two; and both factors will be negative if x 3 6 0, since it’s the larger. The solution set is found by solving these two simple inequalities: x 4 7 0
gives x 7 4 and x 3 6 0 gives x 6 3. If the inequality were 1x 42 1x 32 6 0 instead, we require one negative
factor and one positive factor. Due to order relations and the number line, the larger factor must be the positive one:
x 3 7 0 so x 7 3. The smaller factor must be the negative one: x 4 6 0 and x 6 4. This gives the solution
3 6 x 6 4 as can be verified using any alternative method. Solutions to all other polynomial and rational inequalities
are an extension of these two cases.
Illustration 1 䊳 Solve x 3 7x 6 6 0 using the push principle.
Solution 䊳 The polynomial can be factored using the tests for 1 and 1 and synthetic division. The factors are
1x 22 1x 12 1x 32 6 0, which we’ve conveniently written in increasing order. For the product of three factors to
be negative we require: (1) three negative factors or (2) one negative and two positive factors. The first condition is
met by simply making the largest factor negative, as it will ensure the smaller factors are also negative: x 3 6 0 so
x 6 3. The second condition is met by making the smaller factor negative and the “middle” factor positive:
x 2 6 0 and x 1 7 0. The second solution interval is x 6 2 and x 7 1, or 1 6 x 6 2.
Note the push principle does not require the testing of intervals between the zeroes, nor the “cross/bounce” analysis
at the zeroes and vertical asymptotes (of rational functions). In addition, irreducible quadratic factors can still be ignored
as they contribute nothing to the solution of real inequalities, and factors of even multiplicity can be overlooked precisely
because there is no sign change at these roots.
Illustration 2 䊳 Solve 1x2 12 1x 22 2 1x 32 0 using the push principle.
Solution 䊳 Since the factor 1x2 12 does not affect the solution set, this inequality will have the same solution as
1x 22 2 1x 32 0. Further, since 1x 22 2 will be nonnegative for all x, the original inequality has the same
solution set as 1x 32 0! The solution is x 3.
With some practice, the push principle can be a very effective tool. Use it to solve the following exercises. Check all
solutions by graphing the function on a graphing calculator.
x1
Exercise 1: x3 3x 18 0
Exercise 2: 2
7 0
x 4
Exercise 3: x3 13x 12 6 0
Exercise 4: x3 3x 2 0
Exercise 5: x4 x2 12 7 0
Exercise 6: 1x2 52 1x2 92 1x 22 2 1x 12 0
CUMULATIVE REVIEW CHAPTERS 1–4
1. Solve for R:
1
1
1
R
R1
R2
2
5
1 2
x1
x 1
3. Factor each expression completely:
2. Solve for x:
a. x3 1
b. x3 3x2 4x 12
4. Solve using the quadratic formula. Write answers in
both exact and approximate form: 2x2 4x 1 0.
5. Solve the following inequality: x 3 6 5 or
5 x 6 4.
6. Name the eight toolbox functions, give their
equations, then draw a sketch of each.
7. Use substitution to verify that x 2 3i is a
solution to x2 4x 13 0.
8. Solve the rational inequality:
x4
6 3.
x2
9. As part of a study on traffic conditions, the mayor of
a small city tracks her driving time to work each day
for six months and finds a linear and increasing
relationship. On day 1, her drive time was 17 min.
By day 61 the drive time had increased to 28 min.
Find a linear function that models the drive time and
use it to estimate the drive time on day 121, if the
trend continues. Explain what the slope of the line
means in this context.
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CHAPTER 4 Polynomial and Rational Functions
10. Does the relation shown represent a function? If not,
discuss/explain why not.
Jackie Joyner-Kersee
Sarah McLachlan
Hillary Clinton
Sally Ride
Venus Williams
21. Given Y1 1.47x3 0.51x2 1.99, use your
calculator to evaluate Y1 142, Y1 13.712 , and
Y1 10.092 . Round your answers to two decimal places
if necessary.
Astronaut
Tennis Star
Singer
Politician
Athlete
11. The data given shows the
Exercise 11
profit of a new company
Month Profit (1000s)
for the first 6 months of
5
1
business, and is closely
modeled by the function
13
2
p1m2 1.18x2 10.99x 4.6;
18
3
where p1m2 is the profit
20
4
earned in month m. Assuming
5
21
this trend continues, use
6
19
this function to find the
first month a profit will be
earned 1p 7 02.
1
3 using
12. Graph the function g1x2 1x 22 2
transformations of a basic function.
3
13. Find f 1 1x2, given f 1x2 22x 3, then use
composition to verify your inverse is correct.
14. Graph f 1x2 x2 4x 7 by completing the
square, then state intervals where:
a. f 1x2 0
b. f 1x2c
15. Given the graph of a general
function f 1x2, graph
F1x2 f 1x 12 2.
16. Graph the piecewise-defined
function given:
3
f 1x2 • x
3x
x 6 1
1 x 1
x 7 1
Exercise 15
22. Use a graphing calculator to locate the point(s) of
1.12 3x2
intersection of the graph of r1x2 1.5x2 2.2x
with its horizontal asymptote. Round your answer(s)
to two decimal places if necessary.
23. Use a calculator in complex mode to express the
following complex numbers in a bi form. Round
to the nearest hundredth if necessary.
a. 12.4 1.2i2 6
7 6i
b.
4 5i
c. 10.3 8.2i2 11.9 3.3i2
d. i15
24. Identify intervals containing the zeroes of
f 1x2 x4 3x3 x2 7x 3 by using the
intermediate value theorem and the table feature of
your graphing calculator. Use TblStart 5 and
¢Tbl 0.1, so none of the intervals are greater than
0.1 units wide. Assume all real zeroes are between 5
and 5.
25. Using the calculator screen shown, identify which
function (Y2 through Y7) performs the indicated
transformation of Y1. Verify your answers with your
calculator.
y
5
4
3
2
1
54321
1
2
3
4
5
f(x)
1 2 3 4 5 x
17. Y varies directly with X and inversely with the
square of Z. If Y 10 when X 32 and Z 4, find
X when Z 15 and Y 1.4.
18. Use the rational zeroes theorem and synthetic
division to find all zeroes (real and complex) of
f 1x2 x4 2x2 16x 15.
19. Sketch the graph of f 1x2 x3 3x2 6x 8.
x1
and use the
x2 4
zeroes and vertical asymptotes to solve h1x2 0.
20. Sketch the graph of h1x2 Exercises 21 through 25 require the use of a graphing
calculator.
a.
b.
c.
d.
e.
f.
Y1 shifted up 1 unit
Y1 shifted down 1 unit
Y1 shifted right 1 unit
Y1 shifted left 1 unit
Y1 reflected across x-axis
Y1 reflected across y-axis
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Precalculus—
CONNECTIONS TO CALCULUS
In a calculus course, many of the graphing techniques demonstrated in Chapters 2, 3,
and 4 are applied to a wider variety of functions, but with little change in the basic approach. The graph of a function will always depend on its domain, end-behavior,
y-intercept, x-intercept(s), midinterval points, maximum and minimum values, intervals where the function is positive or negative, and so on. In addition, these elements
remain in play even when the function is not polynomial or rational, as demonstrated
in Example 1. Further, locating the maximum and minimum values of a function is a
vital component of calculus, as in its application we often seek the maximum illumination, lowest cost, greatest efficiency, and the like. These ideas behind locating
extreme values are addressed in Examples 2 and 3.
Graphing Techniques
EXAMPLE 1
Solution
䊳
䊳
Graphing a Radical Function
Graph the function f 1x2 ⫽ x 12 ⫹ x.
y
From the radical factor we note the domain is
x 僆 3 ⫺2, q 2. The y-intercept is (0, 0), with
x-intercepts at (0, 0) and 1⫺2, 02. From the given
expression, we see that as x S q, y S q, and the
end-behavior will be “up on the right.” Evaluating
at the “midinterval point” x ⫽ ⫺1 gives
f 1⫺12 ⫽ ⫺1, showing f 1x2 6 0 in the interval
1⫺2, 02, and f 1x2 7 0 for x 7 0. Using this
information and connecting the given points with a
smooth curve produces the graph shown.
2
1
⫺2
⫺1
1
2
x
⫺1
⫺2
Now try Exercises 1 through 6
䊳
From the graph of f 1x2 ⫽ x 12 ⫹ x in Example 1, the point 1⫺1, ⫺12 appears to be
near the minimum value of the function. However, as we noted in Chapter 4, maximum
and minimum values of a function rarely occur at these “midinterval points,” and finding them often requires the tools of calculus and good algebra skills.
EXAMPLE 2
䊳
Locating Maximum or Minimum Values
Solution
䊳
Substituting 0 for f ¿ 1x2 produces the following sequence:
Using calculus, it can be shown that the minimum value of f 1x2 ⫽ x12 ⫹ x
1
b ⫹ 12 ⫹ x. Find the zero
actually occurs at the zero of f ¿ 1x2 ⫽ xa
2 12 ⫹ x
and locate this minimum value.
x
⫹ 12 ⫹ x
2 12 ⫹ x
x
⫺12 ⫹ x ⫽
2 12 ⫹ x
x2
2⫹x⫽
412 ⫹ x2
412 ⫹ x2 2 ⫽ x2
0⫽
WORTHY OF NOTE
The rotation f ¿1x2 is read “f prime of
x,” and is commonly used to denote
the slope of the line drawn tangent
to the curve at x.
4–99
subtract 12 ⫹ x
square both sides
multiply by 4(2⫹x )
405
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Precalculus—
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4–100
Connections to Calculus
16 ⫹ 16x ⫹ 4x2 ⫽ x2
3x2 ⫹ 16x ⫹ 16 ⫽ 0
13x ⫹ 42 1x ⫹ 42 ⫽ 0
4
x ⫽ ⫺4
x⫽⫺
3
expand binomial square and distribute 4
set equal to zero
factor
result
4
Since ⫺4 is outside the domain of f, the minimum value must occur at x ⫽ ⫺ .
3
Note the zero(s) of f ¿ will tell us where the minimum value occurs, but to find this
minimum we must use this zero in the original function. A minimum value of
4
4
4
4
f a⫺ b ⬇ ⫺1.1 will occur at a⫺ , f a⫺ bb ⬇ a⫺ , ⫺1.1b.
3
3
3
3
Now try Exercises 7 through 10
䊳
In Exercise 86 of Section 4.3, the function model for a population of insects was
given as p1m2 ⫽ ⫺m4 ⫹ 26m3 ⫺ 217m2 ⫹ 588m, where p(m) represents the insect
population (in hundreds of thousands) for month m. As mentioned in the chapter
opener, health officials may be very interested in the month(s) that this population
reaches a peak, as they prepare vaccines or attempt to raise public awareness. A graph
of the function provides a visual representation of population growth, and helps to
identify what months need to be targeted.
EXAMPLE 3
䊳
Locating Maximum and Minimum Values
p(m)
The graph of
600
p1m2 ⫽ ⫺m4 ⫹ 26m3 ⫺ 217m2 ⫹ 588m
500
(modeling the population of insects) is
shown. Using the tools of calculus, it can
be shown that the zeroes of
p¿ 1m2 ⫽ ⫺4m ⫹ 78m ⫺ 434m ⫹ 588
3
䊳
300
2
give the location of the maximum and
minimum values of p:
a. Find these zeroes.
b. Then substitute these zeroes into p(m)
to find the actual maximum and
minimum values.
Solution
400
200
100
1
2
⫺100
3
4
5
6
7
8
9 10 11 12
Months
a. To begin, we set the equation equal to zero and factor out ⫺2:
0 ⫽ ⫺212m3 ⫺ 39m2 ⫹ 217m ⫺ 2942
The tests for 1 and ⫺1 show neither is a zero. Using x ⫽ 2 and the remainder
theorem to help factor p ¿ gives
2
2
2
⫺39
4
⫺35
217
⫺70
147
⫺294
294
0
This shows m ⫽ 2 is a zero and m ⫺ 2 is a factor, resulting in the equation
0 ⫽ ⫺21m ⫺ 22 12m2 ⫺ 35m ⫹ 1472. Using the quadratic formula (or trial21
and-error factoring) shows the remaining zeroes of p ¿ are m ⫽ 7 and m ⫽ .
2
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Connections to Calculus
407
b. By inspecting the graph of p it indeed appears the maximum/minimum values
(peaks and valleys) occur at m ⫽ 2, 7, and 10.5, and it only remains to find the
actual values by substituting these into p(m).
p122 ⫽ ⫺122 4 ⫹ 26122 3 ⫺ 217122 2 ⫹ 588122
⫽ ⫺16 ⫹ 208 ⫺ 868 ⫹ 1176
⫽ 500
p172 ⫽ ⫺172 4 ⫹ 26172 3 ⫺ 217172 2 ⫹ 588172
⫽ ⫺2401 ⫹ 8918 ⫺ 10,633 ⫹ 4116
⫽0
p110.52 ⫽ ⫺110.52 4 ⫹ 26110.52 3 ⫺ 217110.52 2 ⫹ 588110.52
⫽ 192.9375
The maximum infestation of 50,000,000 insects occurs at m ⫽ 2 (February),
and a minimum population of 0 insects occurs at m ⫽ 7 (July). The population
of insects then peaks at a lower level of about 19,300,000 insects at m ⫽ 10.5
(mid-October).
Now try Exercises 11 and 12
䊳
In Examples 2 and 3 it’s important to note that a maximum or minimum value is
an output value (from the range), and not an ordered pair. Considering the “cause and
effect” nature of a function, the cause of a higher or lower insect population is the time
of year, the effect is the population of insects at that time.
Connections to Calculus Exercises
State the domain, end-behavior, and y-intercept of each function. Then find the x-intercepts and use any midinterval
points or other graphing “tools” needed to complete the graph.
1. f 1x2 ⫽ x 13 ⫺ x
2. g1x2 ⫽ 2x2 1x ⫺ 6x 1x
3. h1x2 ⫽ x29 ⫺ x2
3
4. v1x2 ⫽ 1
2x1x ⫹ 5
5. The anxiety level of an Olympic skater is modeled
by the function p 1x2 ⫽ x3 ⫹ 2x2 ⫺ 5x ⫺ 6, where
p(x) is his anxiety level at time x in minutes, and
x ⫽ 0 is the start time of his skating routine
1p 6 0 S calm and relaxed, p 7 0 S anxious,
p ⫽ 0 S normal2. State the domain, end-behavior,
and y-intercept of p, then find the x-intercepts and
use any midinterval points or other graphing
“tools” needed to complete the graph.
6. From a height of 1200 ft, an ASARI plane
(advanced sonar and radar imaging) flies over a
canyon to ascertain the depth of a channel cut by a
river through the canyon. The main point of
interest is the location where a tall hill remains in
the middle of the channel. Using the data gathered,
the operators determine the terrain can be modeled
by the function h 1x2 ⫽ x4 ⫺ 10x2 ⫹ 9, where h(x)
represents the height (in hundreds of feet) at a
distance of x hundred feet from the top of the hill.
State the domain, end-behavior, and y-intercept of
h, then find the x-intercepts and use any
midinterval points or other graphing “tools” needed
to complete the graph.
Using calculus, it can be shown that the maximum or minimum value of the functions from 1 through 6 (designated
by lower case letters), actually occur at the zeroes of the corresponding functions in 7 through 12. Find and clearly
state the maximum or minimum value(s) for each function, and where they will occur. Round to two decimal places
as needed.
7. f ¿ 1x2 ⫽ 13 ⫺ x ⫺
3
2
x
2 13 ⫺ x
9. h¿ 1x2 ⫽
1
2
8. g¿1x2 ⫽ 5x ⫺ 9x
10. v¿ 1x2 ⫽
⫺x2
29 ⫺ x
2
12x2 3
⫹ 29 ⫺ x2
1
1
1
21x ⫹ 52 2
⫹
21x ⫹ 52 2
2
312x2 3
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Connections to Calculus
11. Using the tools of calculus, it can be shown
that the maximum and minimum values of p
(Exercise 5) occur at the zeroes of
p¿1x2 ⫽ 3x2 ⫹ 4x ⫺ 5. Determine the maximum
level of anxiety felt by the skater and when it
occurs. What is the minimum level of anxiety?
When does it occur? (Note: In realistic terms
we might consider the domain of p to be
⫺3 ⱕ x ⱕ 3.2
4–102
12. Using the tools of calculus, it can be shown that the
maximum and minimum values of h (Exercise 6)
occur at the zeroes of h¿1x2 ⫽ 4x3 ⫺ 20x.
Determine the maximum height, and where this
maximum occurs, then find the depth of the first
channel the plane flies over. How far from the top
of the hill does this occur? What do you notice
about the depth of the second channel? (Note: In
realistic terms we might consider the domain of h
to be about ⫺3.1 ⱕ x ⱕ 3.1.)